- 最长连续序列
- 3SUM
- 923. 三数之和的多种可能
- 300. 最长上升子序列
- 333. 最大 BST 子树
- 33. 搜索旋转排序数组
- 153. 寻找旋转排序数组中的最小值
- 545. 二叉树的边界
- 679. 24点游戏
- 312. 戳气球
- 1246. 删除回文子数组
- 772. 基本计算器 III
- 727. 最小窗口子序列
- 642. 设计搜索自动补全系统
- 702. 搜索长度未知的有序数组
- 146. LRU缓存机制
- 490. 迷宫
- 340. 至多包含 K 个不同字符的最长子串
- 240. 搜索二维矩阵 II
- 428. 序列化和反序列化 N 叉树
- 253. 会议室 II
- 1236. 网络爬虫
- 727. 最小窗口子序列
- 694. 不同岛屿的数量
- 295. 数据流的中位数
- 140. 单词拆分 II
- 138. 复制带随机指针的链表
- 510. 二叉搜索树中的中序后继 II
- 545. 二叉树的边界
- 314. 二叉树的垂直遍历
- 722. 删除注释
最长连续序列
class Solution{
public:
int longestConsecutive(vector<int> &nums){
unordered_map<int , int> map;
int size = nums.size();
int l = 1;
for(int i = 0; i < size ; i++){
if(map.find(nums[i] != map.end())){
continue;
}
map[nums[i]] = 1;
if(map.find(nums[i]-1) != map.end()){
l = max(l, mergeCluster(map, nums[i] - 1, nums[i]));
}
if(map.find(nums[i]+1) != map.end()){
l = max(l, mergeCluster(map, nums[i], nums[i] + 1));
}
}
return size == 0 ? 0 : l;
}
private:
int mergeCluster(unordered_map<int, int>& map, int left ,int right){
int upper = right + map[right] - 1;
int lower = left - map[left] + 1;
int leghth = upper - lower + 1;
map[upper] = length;
map[lower] = length;
return length;
}
}
3SUM
class So;ution{
public:
vetcor<vector<int>> threeSum(vector<int>& nums, const int target){
vector<vector<int>> result;
if(nums.size() < 3) return result;
sort(nums.begin(), nums.end());
auto last = nums.end();
for(auto i = nums.begin(); i < last - 2; i++){
auto j = i + 1;
if(j > nums.beigin() && *i == *(i-1)){
continue;
}
auto k = last - 1;
while(j < k){
if(*i + *j + *k < target){
++j;
while(*j == *(j+1) && j < k){
j++;
}
}else if(*i + *j + *k > target){
k--;
while(*k == *(k+1) && j < k){
k--;
}
}else{
result.push_back({*i, *j ,*k});
j++;
k--;
while(*j == *(j-1) )
}
}
}
}
}
923. 三数之和的多种可能
class Solution {
public:
const long D = 1e9 + 7;
int threeSumMulti(vector<int>& nums, int target) {
int res = 0;
if(nums.size() < 3) return res;
int size = nums.size();
sort(nums.begin(), nums.end());
for(auto i = 0; i < size - 2; ++i){
if(target < 3 * nums[i]) break;
int j = i + 1;
int k = size - 1;
while(nums[j] < nums[k]){
if (nums[i] + nums[j] + nums[k] < target){
++j;
} else if(nums[i] + nums[j] + nums[k] > target){
--k;
} else {
int tmpJ = j;
int sameJ = 1;
int sameK = 1;
while(j + 1 <= k && nums[tmpJ] == nums[++j]){//不用tmpJ的写法可能会死循环，为什么？
sameJ++;
}
int tmpK = k;
while(k - 1 >= j && nums[tmpK] == nums[--k]){
sameK++;
}
res += (sameJ * sameK);
res = res % D;
}
}
if(nums[j] == nums[k] && nums[i] + nums[j] + nums[k] == target){
long d = k -j + 1;
res += (d * (d-1)) / 2;
res = res%D;
}
}
return res%D;
}
};
300. 最长上升子序列
长度为 i(下标) + 1 的递增子序列末尾的最小数字。
很具小巧思。新建数组 cell，用于保存最长上升子序列。
对原序列进行遍历，将每位元素二分插入 cell 中。
如果 cell 中元素都比它小，将它插到最后
否则，用它覆盖掉比它大的元素中最小的那个。
尝试left + 1 < right 模板
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
if(nums.empty()) return 0;
if(nums.size() == 1) return 1;
int len = nums.size();
vector<int> dp(len , 0);
int index = 0;
dp[index] = nums[0];
for(int i = 1; i < len; i++){
if(nums[i] > dp[index]){
dp[++index] = nums[i];
} else {
int l = 0;
int r = index;
while(l + 1 < r){
int mid = l + (r - l)/2;
if(nums[i] >= dp[mid] ){//nums[i] <= dp[mid] 也行
l = mid;
} else if(nums[i] < dp[mid]){
r = mid;
}
}
if (dp[l] > nums[i] && nums[i] != dp[r]){
dp[l] = nums[i];
} else if(dp[r] > nums[i] && dp[l] != nums[i]){
dp[r] = nums[i];
}
}
}
return index + 1;
}
};
333. 最大 BST 子树
helper函数返回了一个一维数组，里面有三个数字，分别是以当前结点为根结点的树的最小值，最大值，以及最大的 BST 子树的结点个数。
class Solution {
public:
int largestBSTSubtree(TreeNode* root) {
vector<int> res = help(root);
return res[2];
}
vector<int> help(TreeNode* node){
if(node == nullptr){
return {INT_MAX, INT_MIN, 0};
}
vector<int> left = help(node->left);
vector<int> right = help(node->right);
if(node->val > left[1] && node->val < right[0]){
return {min(node->val, left[0]), max(node->val, right[1]), left[2] + right[2] + 1};
}else{
return {INT_MIN, INT_MAX, max(left[2], right[2])};
}
}
};
33. 搜索旋转排序数组
先锁定mid的区间
class Solution {
public:
int search(vector<int>& nums, int target) {
if(nums.empty()){
return -1;
}
int len = nums.size();
int l = 0;
int r = nums.size() - 1;
while(l + 1 < r){
int mid = l + (r - l)/2;
if(nums[mid] == target) return mid;
if(nums[0] <= nums[mid]){
if(nums[0] <= target && target <= nums[mid]){
r = mid;
} else {
l = mid;
}
} else {
if(nums[len - 1] >=target && target >= nums[mid]){
l = mid;
}else{
r = mid;
}
}
}
if(nums[l] == target){
return l;
}else if(nums[r] == target){
return r;
}else{
return -1;
}
}
};
153. 寻找旋转排序数组中的最小值
class Solution {
public:
int findMin(vector<int>& nums) {
if(nums.empty()){
return -1;
}
int len = nums.size();
int l = 0;
int r = len - 1;
while(l + 1 < r){
int mid = l + (r - l)/2;
if(nums[0] > nums[len-1]){
if(nums[mid] >= nums[0]){
l = mid;
}
else if(nums[mid] <= nums[len -1 ]){
r = mid;
}
}else{
return nums[0]; // 注意不旋转的情况
}
}
return min(nums[l], nums[r]);
}
};
545. 二叉树的边界
class Solution {
public:
void dfs(TreeNode* root, int dir, vector<int>& res){
if(root == nullptr) return;
if(dir == 0){
//无方向，寻找叶子节点
if(root->left == nullptr && root->right == nullptr){
res.push_back(root->val);
}else{
dfs(root->left, 0, res);
dfs(root->right, 0, res);
}
return;
}
if(dir == -1){
//左方向，先序遍历
res.push_back(root->val);
if(root->left){
dfs(root->left, dir, res);
dfs(root->right, 0, res);
}else{
dfs(root->right, dir, res);
}
}else{
//右方向，后序遍历
if(root->right){
dfs(root->left, 0, res);
dfs(root->right, dir, res);
}else{
dfs(root->left, dir, res);
}
res.push_back(root->val);
}
}
vector<int> boundaryOfBinaryTree(TreeNode* root) {
if(root == nullptr) return{};
vector<int> res{root->val};
dfs(root->left, -1, res);
dfs(root->right, 1, res);
return res;
}
};
679. 24点游戏
class Solution {
public:
vector<char> ops{'+', '-', '*', '/'};
double eps = 0.001;
bool judgePoint24(vector<int>& nums) {
vector<double> arr(nums.begin(), nums.end());
return helper(arr);
}
bool helper(vector<double>& nums) {
if (nums.size() == 1) return abs(nums[0] - 24) < eps;
for (int i = 0; i < nums.size(); ++i) {
for (int j = 0; j < nums.size(); ++j) {
if (i == j) continue;
vector<double> t;
for (int k = 0; k < nums.size(); ++k) {
if (k != i && k != j) t.push_back(nums[k]);
}
for (char op : ops) {
if ((op == '+' || op == '*') && i > j) continue;
if (op == '/' && nums[j] < eps) continue;
switch(op) {
case '+': t.push_back(nums[i] + nums[j]); break;
case '-': t.push_back(nums[i] - nums[j]); break;
case '*': t.push_back(nums[i] * nums[j]); break;
case '/': t.push_back(nums[i] / nums[j]); break;
}
if (helper(t)) return true;
t.pop_back();
}
}
}
return false;
}
};
312. 戳气球
https://leetcode-cn.com/problems/burst-balloons/
class Solution {
public:
int maxCoins(vector<int>& nums) {
nums.insert(nums.begin(),1);
nums.push_back(1);
int n=nums.size();
vector<vector<int>> dp(n, vector<int>(n, 0));//dp[i][j]表示第i至第j个元素这个区间能获得的最大硬币数
for(int r=2;r<n;r++) //r为区间长度
for(int i=0;i<n-r;i++){ //i为左区间
int j=i+r; //j为右区间
for(int k=i+1;k<j;k++)
dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]+nums[i]*nums[k]*nums[j]);
}
return dp[0][n-1];
}
};
1246. 删除回文子数组
class Solution {
public:
int minimumMoves(vector<int>& arr) {
int dp[arr.size()][arr.size()];
for(int i = 0; i < arr.size(); i++)
dp[i][i] = 1;
for(int i = 0; i < arr.size() - 1; i++)
if(arr[i] == arr[i + 1])
dp[i][i + 1] = 1;
else
dp[i][i + 1] = 2;
for(int len = 2; len < arr.size(); len++){
for(int i = 0; i < arr.size() - len; i++) {
dp[i][i + len] = len + 1;
for(int k = 0; k < len; k++){
if(dp[i][i + len] > dp[i][i + k] + dp[i + k + 1][i + len]){
dp[i][i + len] = dp[i][i + k] + dp[i + k + 1][i + len];
}
}
if(arr[i] == arr[i + len] && dp[i][i + len] > dp[i + 1][i + len - 1]){
dp[i][i + len] = dp[i + 1][i + len - 1];
}
}
}
return dp[0][arr.size() - 1];
}
};
772. 基本计算器 III
class Solution {
public:
int calculate(string s) {
}
};
727. 最小窗口子序列
class Solution {
public:
string minWindow(string S, string T) {
if (S.size() < T.size()) return "";
int NS = S.size();
int NT = T.size();
string res;
int resLen = NS;
for (int i = 0; i <= NS - NT; ++i) {
if (S[i] != T[0]) continue;
// 正向匹配
int j = 0;
while (i < NS && j < NT) {
if (S[i] == T[j]) ++j;
++i;
}
if (j != NT) break;
// 反向匹配
int r = i;
j = NT - 1;
--i;
while (j >= 0) {
if (S[i] == T[j]) --j;
--i;
}
++i;
if (r - i < resLen) {
res = S.substr(i, r - i);
resLen = r - i;
} else {
i = r - resLen;
}
}
return res;
}
};
642. 设计搜索自动补全系统
struct TrieNode {
int val = 0;
map<char, TrieNode*> children;
};
class AutocompleteSystem {
public:
AutocompleteSystem(vector<string>& sentences, vector<int>& times) {
ans = "";
cur = root = new TrieNode();
// 构建trie树
for(int i=0;i<sentences.size();i++) {
add(sentences[i], times[i]);
}
}
void add(string& s, int time) {
TrieNode* node = root;
for(char c: s) {
if(node->children.find(c)==node->children.end()) {
node->children[c] = new TrieNode();
}
node = node->children[c];
}
node->val += time;
}
vector<string> input(char c) {
// 输入结束时复位
if (c == '#') {
cur->val++;
cur = root;
ans = "";
return {};
}
// 更新当前节点
if(cur->children.find(c)==cur->children.end()) {
cur->children[c] = new TrieNode();
}
cur = cur->children[c];
ans += c;
// 查找所有符合条件的候选集及出现次数
vector<pair<string, int>> vec;
find(cur, ans, vec);
// 按次数及字典序排序
sort(vec.begin(), vec.end(), [this](pair<string, int>& p1, pair<string, int>& p2){
return p1.second == p2.second ? compare(p1.first, p2.first) : p1.second > p2.second;
});
// 取top 3
vector<string> res;
for(auto& p: vec) {
res.push_back(p.first);
if(res.size()>=3) break;
}
return res;
}
// 字典序比较
bool compare(string& s1, string& s2) {
int i = 0, m = s1.size(), n = s2.size();
while(i<m&&i<n) {
if(s1[i] != s2[i]) {
return s1[i] < s2[i];
}
i++;
}
return m <= n;
}
// dfs查找所有候选集及次数
void find(TrieNode* node, string ans, vector<pair<string, int>>& vec) {
if(node->val>0) {
vec.push_back({ans, node->val});
}
for(auto& it: node->children) {
find(it.second, ans+it.first, vec);
}
}
private:
TrieNode* root;
TrieNode* cur; // 当前节点
string ans = ""; // 历史输入
};
702. 搜索长度未知的有序数组
class ArrayReader;
class Solution {
public:
int search(const ArrayReader& reader, int target) {
if (reader.get(0) == target) return 0;
// search boundaries
int left = 0, right = 1;
while (reader.get(right) < target) {
left = right;
right <<= 1;
}
// binary search
int pivot, num;
while (left <= right) {
pivot = left + ((right - left) >> 1);
num = reader.get(pivot);
if (num == target) return pivot;
if (num > target) right = pivot - 1;
else left = pivot + 1;
}
// there is no target element
return -1;
}
};
146. LRU缓存机制
class LRUCache {
private:
int cap;
// 双链表：装着 (key, value) 元组
list<pair<int, int>> cache;
// 哈希表：key 映射到 (key, value) 在 cache 中的位置
unordered_map<int, list<pair<int, int>>::iterator> map;
public:
LRUCache(int capacity) {
this->cap = capacity;
}
int get(int key) {
auto it = map.find(key);
// 访问的 key 不存在
if (it == map.end()) return -1;
// key 存在，把 (k, v) 换到队头
pair<int, int> kv = *map[key];
cache.erase(map[key]);
cache.push_front(kv);
// 更新 (key, value) 在 cache 中的位置
map[key] = cache.begin();
return kv.second; // value
}
void put(int key, int value) {
/* 要先判断 key 是否已经存在 */
auto it = map.find(key);
if (it == map.end()) {
/* key 不存在，判断 cache 是否已满 */
if (cache.size() == cap) {
// cache 已满，删除尾部的键值对腾位置
// cache 和 map 中的数据都要删除
auto lastPair = cache.back();
int lastKey = lastPair.first;
map.erase(lastKey);
cache.pop_back();
}
// cache 没满，可以直接添加
cache.push_front(make_pair(key, value));
map[key] = cache.begin();
} else {
/* key 存在，更改 value 并换到队头 */
cache.erase(map[key]);
cache.push_front(make_pair(key, value));
map[key] = cache.begin();
}
}
};
490. 迷宫
class Solution {
public:
bool BFS(vector<vector<int>>& maze, vector<vector<bool>> &visit, vector<int> start, vector<int> end, int matrix_r, int matrix_c) {
queue<pair<int, int>> q;
// 队首元素入队
q.push(make_pair(start[0], start[1]));
visit[start[0]][start[1]] = true;
while (!q.empty()) {
pair<int, int> top = q.front();
q.pop();
// 取队列首元素，确定下是否到达终点
if (top.first == end[0] && top.second == end[1]) {
return true;
}
int x[4] = {0, 0, 1, -1};
int y[4] = {1, -1, 0, 0};
for (int i = 0; i < 4; i++) {
int row = top.first;
int col = top.second;
// 球回沿着一个方向一直滚动， 除非遇到墙才会停下来选择别的方向
while ((row + x[i]) >= 0 && (row + x[i]) < matrix_r && (col + y[i]) >= 0 && (col +y[i]) < matrix_c &&
maze[row + x[i]][col + y[i]] == 0) {
row += x[i];
col += y[i];
}
// 等到碰到墙壁，需要更换方向前，需要将此点入队，并标志已经访问
// 队列中存放的是碰到墙壁的点或者边界点（可以理解为墙壁点）
if (visit[row][col] == false) {
q.push(make_pair(row, col));
visit[row][col] = true;
}
}
}
return false;
}
bool hasPath(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
int matrix_r = maze.size();
int matrix_c = maze[0].size();
vector<vector<bool>> visit(matrix_r, vector<bool>(matrix_c, false));
bool res = BFS(maze, visit , start, destination, matrix_r, matrix_c);
return res;
}
};
340. 至多包含 K 个不同字符的最长子串
class Solution {
public:
int lengthOfLongestSubstringKDistinct(string s, int k) {
int cnt[256] = {0};
int used = 0;
int len = 0, start = 0;
for(int i = 0; i < s.length(); i++)
{
if(0 == cnt[s[i]]) used++;
cnt[s[i]]++;
while(used > k)
{
cnt[s[start]]--;
if(0 == cnt[s[start]]) used--;
start++;
}
len = max(len, i - start + 1);
}
return len;
}
};
240. 搜索二维矩阵 II
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size() == 0 || matrix[0].size() == 0) return false;
int i = matrix.size() - 1;
int j = 0;
while(i >= 0 && j < matrix[0].size()){
if(matrix[i][j] > target)
--i;
else if(matrix[i][j] < target)
++j;
else return true;
}
return false;
}
};
428. 序列化和反序列化 N 叉树
class Codec {
public:
// Encodes a tree to a single string.
string serialize(Node* root) {
if (root == NULL)
return "";
string res = to_string(root->val) + "[";
for (auto node : root->children) {
res += serialize(node) + ",";
}
if (res.back() == ',')
res.pop_back();
res += "]";
return res;
}
// Decodes your encoded data to tree.
Node* deserialize(string data) {
cout << data << endl;
string value;
Node* head = NULL;
stack<Node*> s;
for (int i = 0; i < data.size(); ++i) {
const char& c = data[i];
if ((c >= '0' && c <= '9') || c == '+' || c == '-') {
value += c;
} else if (c == '[') {
Node* node = new Node(stoi(value));
if (head == NULL)
head = node;
s.push(node);
value.clear();
} else if (c == ']') {
Node* node = s.top();
s.pop();
if (!s.empty()) {
Node* prev_node = s.top();
prev_node->children.push_back(node);
}
}
}
return head;
}
};
253. 会议室 II
A priority queue is a container adaptor that provides constant time lookup of the largest (by default) element, at the expense of logarithmic insertion and extraction.
默认是大顶堆，本题使用小顶堆
结束时间最小的是否满足接下来要开始的
class Solution {
public:
int minMeetingRooms(vector<vector<int>>& intervals) {
if (intervals.size() == 0)
return 0;
// 根据会议开始时间排序
sort(intervals.begin(), intervals.end());
priority_queue<int, vector<int>, greater<int>> rooms;
for (const auto& range : intervals) {
const int& s = range[0];
const int& e = range[1];
if (rooms.empty()) { // 第一个会议结束时间来初始堆
rooms.push(e);
continue;
} else if (rooms.top() <= s) { // 有空余房间
rooms.pop();
rooms.push(e);
} else rooms.push(e); // 没有空余房间
}
return rooms.size(); // 返回房间个数
}
};
1236. 网络爬虫
class Solution {
public:
string getHostname(string& url) {
int pos = url.find('/', 7);
if (pos == string::npos) {
return url.substr(7);
}
return url.substr(7, pos - 7);
}
vector<string> crawl(string startUrl, HtmlParser htmlParser) {
set<string> visited;
deque<string> q;
q.push_back(startUrl);
vector<string> res;
string hostname = getHostname(startUrl);
while (!q.empty()) {
string url = q.front();
q.pop_front();
if (visited.find(url) != visited.end()) {
continue;
}
visited.insert(url);
if (getHostname(url) != hostname) {
continue;
}
res.push_back(url);
vector<string> urls = htmlParser.getUrls(url);
for (string& s : urls) {
q.push_back(s);
}
}
return res;
}
};
727. 最小窗口子序列
1，从S的起点i先正向匹配找到符合条件的区间最小的终点j
2，从j反向匹配，找到最大的满足条件的新起点k
class Solution {
public:
string minWindow(string S, string T) {
if (S.size() < T.size()) return "";
int NS = S.size();
int NT = T.size();
string res;
int resLen = NS;
for (int i = 0; i <= NS - NT; ++i) {
if (S[i] != T[0]) continue;
// 正向匹配
int j = 0;
while (i < NS && j < NT) {
if (S[i] == T[j]) ++j;
++i;
}
if (j != NT) break;
// 反向匹配
int r = i;
j = NT - 1;
--i;
while (j >= 0) {
if (S[i] == T[j]) --j;
--i;
}
++i;
if (r - i < resLen) {
res = S.substr(i, r - i);
resLen = r - i;
} else {
i = r - resLen;
}
}
return res;
}
};
694. 不同岛屿的数量
class Solution {
public:
int row, col;
void dfs(vector<int>& island, vector<vector<bool>>& visited, vector<vector<int>>& grid, int sr, int sc, int r,int c){
if(r>=row || r<0 || c<0 || c>=col || visited[r][c] || grid[r][c] ==0) {
return;
}
visited[r][c] = true;
island.push_back((r-sr)*col + (c-sc));
dfs(island, visited, grid, sr, sc, r-1, c);
dfs(island, visited, grid, sr, sc, r+1, c);
dfs(island, visited, grid, sr, sc, r, c-1);
dfs(island, visited, grid, sr, sc, r, c+1);
}
int numDistinctIslands(vector<vector<int>>& grid) {
row = grid.size();
col = grid[0].size();
vector<vector<bool>> visited(row, vector<bool>(col,false));
set<vector<int>> islands;
for(int i=0;i<row;i++){
for(int j=0;j<col;j++){
vector<int> island;
dfs(island, visited, grid, i, j, i, j);
if(!island.empty()){
islands.insert(island);
}
}
}
return islands.size();
}
};
295. 数据流的中位数
class MedianFinder {
priority_queue<int> lo; // max heap
priority_queue<int, vector<int>, greater<int>> hi; // min heap
public:
// Adds a number into the data structure.
void addNum(int num)
{
lo.push(num); // Add to max heap
hi.push(lo.top()); // balancing step
lo.pop();
if (lo.size() < hi.size()) { // maintain size property
lo.push(hi.top());
hi.pop();
}
}
// Returns the median of current data stream
double findMedian()
{
return lo.size() > hi.size() ? (double) lo.top() : (lo.top() + hi.top()) * 0.5;
}
};
140. 单词拆分 II
class Solution {
public:
vector<string> wordBreak(string s, vector<string>& wordDict) {
unordered_map<string,vector<string> > m;
return helper(m,wordDict,s);
}
vector<string> helper(unordered_map<string,vector<string> >& m,vector<string>& wordDict,string s){
if(m.count(s)) return m[s];
if(s.empty()) return {""};
vector<string> res;
for(auto word : wordDict){
if(s.substr(0,word.size())!=word) continue;
vector<string> tmp=helper(m,wordDict,s.substr(word.size()));
for(auto itm:tmp){
res.push_back(word+(itm.empty()?"":" "+itm));
}
}
m[s]=res;
return res;
}
};
138. 复制带随机指针的链表
class Solution {
public:
//方法1
Node* copyRandomList1(Node* head)
{
if (head == nullptr)
return head;
//遍历原链表 创建新链表节点并建立映射关系
unordered_map<Node*, Node*> map; //key: 原链表节点 value: 新创建节点
Node* cur = head;
while (cur)
{
map[cur] = new Node(cur->val);
cur = cur->next;
}
//遍历原链表 根据map找新链表的random
cur = head;
while (cur)
{
map[cur]->next = map[cur->next];
map[cur]->random = map[cur->random];
cur = cur->next;
}
return map[head];
}
//方法2
Node* copyRandomList(Node* head)
{
if (head == nullptr)
return head;
//遍历原链表 遍历过程中插入新副本节点
Node* cur = head;
while (cur)
{
Node* node = new Node(cur->val);
Node* next = cur->next;
node->next = next;
cur->next = node;
cur = next;
}
//遍历原链表 对新副本节点设置random指针
cur = head;
while (cur)
{
cur->next->random = cur->random ? cur->random->next : nullptr;
cur = cur->next->next;
}
//分离出原链表与新副本链表
cur = head;
Node* new_cur = head->next;
Node* res = new_cur;
while (cur)
{
cur->next = cur->next->next;
cur = cur->next;
new_cur->next = cur ? cur->next : nullptr;
new_cur = new_cur->next;
}
return res; //注意：不能再返回head->next了，head已经是分离后的原链表
}
};
510. 二叉搜索树中的中序后继 II
class Solution {
public:
Node* inorderSuccessor(Node* node) {
if (!node) return node;
if (!node->right) {
Node* head = node->parent;
while (head) {
if (head->val > node->val) break;
head = head->parent;
}
return head;
}
Node* head = node->right;
while (head->left) {
head = head->left;
}
return head;
}
};
545. 二叉树的边界
class Solution {
public:
void dfs(TreeNode* root, int dir, vector<int>& res){
if(root == nullptr) return;
if(dir == 0){
//无方向，寻找叶子节点
if(root->left == nullptr && root->right == nullptr){
res.push_back(root->val);
}else{
dfs(root->left, 0, res);
dfs(root->right, 0, res);
}
return;
}
if(dir == -1){
//左方向，先序遍历
res.push_back(root->val);
if(root->left){
dfs(root->left, dir, res);
dfs(root->right, 0, res);
}else{
dfs(root->right, dir, res);
}
}else{
//右方向，后序遍历
if(root->right){
dfs(root->left, 0, res);
dfs(root->right, dir, res);
}else{
dfs(root->left, dir, res);
}
res.push_back(root->val);
}
}
vector<int> boundaryOfBinaryTree(TreeNode* root) {
if(root == nullptr) return{};
vector<int> res{root->val};
dfs(root->left, -1, res);
dfs(root->right, 1, res);
return res;
}
};
314. 二叉树的垂直遍历
class Solution {
private:
map<int,int> hasht;
vector<vector<int>> ans;
public:
vector<vector<int>> verticalOrder(TreeNode* root) {
if(root == NULL) return ans;
queue<TreeNode*> q;
queue<int> state;
q.push(root);
state.push(0);
while(q.size()!=0){
auto temp = q.front();
auto temp_state = state.front();
q.pop();
state.pop();
if(hasht.find(temp_state) == hasht.end()){
vector<int> ans_layer;
ans_layer.push_back(temp->val);
ans.push_back(ans_layer);
hasht[temp_state] = ans.size()-1;
}
else{
ans[hasht[temp_state]].push_back(temp->val);
}
if(temp->left != NULL){
q.push(temp->left);
state.push(temp_state - 1);
}
if(temp->right != NULL){
q.push(temp->right);
state.push(temp_state + 1);
}
}
vector<vector<int>> ordered_ans;
for(auto &it:hasht){
ordered_ans.push_back(ans[(it).second]);
}
return ordered_ans;
}
};
722. 删除注释
class Solution {
private:
map<int,int> hasht;
vector<vector<int>> ans;
public:
vector<vector<int>> verticalOrder(TreeNode* root) {
if(root == NULL) return ans;
queue<TreeNode*> q;
queue<int> state;
q.push(root);
state.push(0);
while(q.size()!=0){
auto temp = q.front();
auto temp_state = state.front();
q.pop();
state.pop();
if(hasht.find(temp_state) == hasht.end()){
vector<int> ans_layer;
ans_layer.push_back(temp->val);
ans.push_back(ans_layer);
hasht[temp_state] = ans.size()-1;
}
else{
ans[hasht[temp_state]].push_back(temp->val);
}
if(temp->left != NULL){
q.push(temp->left);
state.push(temp_state - 1);
}
if(temp->right != NULL){
q.push(temp->right);
state.push(temp_state + 1);
}
}
vector<vector<int>> ordered_ans;
for(auto &it:hasht){
ordered_ans.push_back(ans[(it).second]);
}
return ordered_ans;
}
};
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