92. 反转链表 II
思路
- 两个指针a、b，分别找到被反转的第一个结点的前一个结点，被反转的结点的最后一个结点，（在开头设置一个哑结点，防止被反转的第一个结点是头结点）
- 再来一个指针c，保存被反转的最后一个结点的next，然后把最后一个结点的next设为null
- 反转链表，然后把新链表的head接回去，把c接回到末尾
- 返回哑结点的next，不能返回head，因为反转以后，head有可能不是head了
AC代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode* dummy = new ListNode(0), *a, *b, *c;
dummy->next = head;
a = b = c = dummy;
for (int i = 0; i < m - 1; i++) {
a = a->next;
b = b->next;
}
for (int i = 0; i < n - m + 1; i++) {
b = b->next;
}
c = b->next;
b->next = NULL;
a->next = reverseList(a->next);
while (a->next != NULL) {
a = a->next;
}
a->next = c;
return dummy->next;
}
ListNode* reverseList(ListNode* head) {
if (head == NULL || head->next == NULL) return head;
ListNode *temp = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return temp;
}
};
15. 三数之和
思路
- 遍历数组，取每个值的相反数作为target，然后转化为两数之和的问题，去重时要注意
- 保证target只查找一次
- 保证第二个循环j = i + 1开始
- 保证查找到的数的下标 c > j
- 保证第二次循环的相同元素对应的值不会被反复查找，即变量find
AC代码
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
int len = nums.size();
vector<vector<int>> ans;
map<int, int> m;
for (int i = 0; i < len; i++) {
m[nums[i]] = i + 1;
}
for (int i = 0; i < len; ) {
int target = -nums[i];
for (int j = i + 1; j < len;) {
int find = target - nums[j];
if (m.count(find)) {
int c = m[find] - 1;
//cout << nums[i] << nums[j] << nums[c] << endl;
if (c > j) {
ans.push_back({nums[i], nums[j], nums[c]});
}
}
while (j < len && nums[j] == target - find) {
j++;
}
}
while (i < len && nums[i] == -target) {
i++;
}
}
return ans;
}
};
大佬思路
二分查找
大佬代码
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
set<vector<int>> ans;
if(nums.size()<3)return vector<vector<int>>(ans.begin(),ans.end());
sort(nums.begin(),nums.end());
int left,right,target;
for(int i=0;i<nums.size()-2;++i){
if(nums[i] > 0)
{
break;
}
if(nums[i] == nums[i - 1] && i > 0)
continue;
left=i+1,right=nums.size()-1,target=-nums[i];
while(left<right){
if(nums[left]+nums[right]==target){
ans.insert({nums[i], nums[left], nums[right]});
++left;
--right;
}else if(nums[left]+nums[right]>target){
--right;
}else {
++left;
}
}
}
return vector<vector<int>>(ans.begin(),ans.end());
}
};
43. 字符串相乘
思路
两层for循环相乘，把相乘的结果全都加起来
AC代码
class Solution {
public:
string multiply(string num1, string num2) {
int len1 = num1.length(), len2 = num2.length();
int len = len1 > len2 ? len1 : len2;
string zero(len - len1, '0');
num1 = zero + num1;
zero = string(len - len2, '0');
num2 = zero + num2;
string ans = "0";
for (int i = len - 1; i >= 0; i--) {
string temp = num1;
int carry = 0;
zero = string(len - 1 - i, '0');
for (int j = len - 1; j >= 0; j--) {
temp[j] = ((num1[j] - '0')*(num2[i] - '0') + carry)%10+ '0';
carry = ((num1[j] - '0')*(num2[i] - '0') + carry)/10;
}
temp = string(1, carry + '0') + temp + zero;
ans = addStrings(ans, temp);
}
int i = 0;
while (ans[i] == '0') i++;
len = ans.length();
return i == len ? "0" : ans.substr(i, len - i);
}
string addStrings(string& num1, string& num2) {
int carry = 0;
int len1 = num1.length();
int len2 = num2.length();
int len = len1 > len2 ? len1 : len2;
string zero = string(len - len1, '0');
num1 = zero + num1;
zero = string(len - len2, '0');
num2 = zero + num2;
for (int i = len - 1; i >= 0; i--) {
int n = carry + num1[i] + num2[i] - 2*'0';
num1[i] = n % 10 + '0';
carry = n / 10;
}
if (carry) num1.insert(0, 1, carry + '0');
return num1;
}
};
73. 矩阵置零
思路
想写出来很简单，目前是空间O(M+N)的算法
AC代码
class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
int r = matrix.size(), c = matrix[0].size();
unordered_map<int, bool> rows, cols;
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (matrix[i][j] == 0) {
rows[i] = true;
cols[j] = true;
}
}
}
for (auto x : rows) {
for (int i = 0; i < c; i++) {
matrix[x.first][i] = 0;
}
}
for (auto x : cols) {
for (int i = 0; i < r; i++) {
matrix[i][x.first] = 0;
}
}
}
};
60. 第k个排列
思路
没研究这个，stl直接调用
AC代码
class Solution {
public:
string getPermutation(int n, int k) {
string ans;
for (int i = 1; i <= n; i++) {
ans += char(i + '0');
}
for (int i = 0; i < k - 1; i++) {
next_permutation(ans.begin(), ans.end());
}
return ans;
}
};
大佬代码
static const auto io_sync_off = []()
{
// turn off sync
std::ios::sync_with_stdio(false);
// untie in/out streams
std::cin.tie(nullptr);
return nullptr;
}();
class Solution {
public:
string recursive(int n, int k, int * order, string &str) {
if (n == 0)
return "";
int num = (k - 1) / order[n - 1];
char c = str[num];
str.erase(str.begin() + num);
return c + recursive(n - 1, k - num * order[n - 1], order, str);
}
string getPermutation(int n, int k) {
int order[n + 1] = {1};
string str;
for (int i = 1; i < n + 1; i++) {
order[i] = i * order[i - 1];
str.push_back(48 + i);
}
return recursive(n, k, order, str);
}
};
34. 在排序数组中查找元素的第一个和最后一个位置
思路
一次二分查找，然后向前向后遍历，找到开始和结束，但是最坏情况下，算法从变成
AC代码
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int len = nums.size();
if (!len) return {-1, -1};
int low = 0, high = len - 1;
bool find = false;
int pos = 0;
while (low <= high) {
int mid = low + (high - low)/2;
if (nums[mid] == target) {
find = true;
pos = mid;
break;
} else if (nums[mid] > target) {
high = mid - 1;
} else {
low = mid + 1;
}
}
if (!find) {
return {-1, -1};
}
int beg , end;
beg = end = pos;
while (beg >= 0 && nums[pos] == nums[beg]) beg--;
while (end < len && nums[pos] == nums[end]) end++;
return {beg + 1, end - 1};
}
};
24. 两两交换链表中的节点
AC代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode dummy(0), *h;
dummy.next = head;//哑结点定义为局部变量，防止内存泄漏
h = &dummy;
while (h->next != NULL) {
if (h->next->next != NULL) {
ListNode *a = h->next, *b = h->next->next;
a->next = b->next;
b->next = a;
h->next = b;
h = h->next->next;
} else {
break;
}
}
return dummy.next;
}
};
47. 全排列 II
AC代码
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> ans;
do {
ans.push_back(nums);
}while (next_permutation(nums.begin(), nums.end()));
ans.erase(unique(ans.begin(), ans.end()),ans.end());
return ans;
}
};
大佬代码
class Solution {
public:
vector<vector<int>> ans;
vector<bool> b;
vector<int> v;
void dfs(int i, const vector<int>& nums)
{
if(i == nums.size()){
ans.push_back(v);
return;
}
for(int j = 0; j < nums.size(); ++j){
if(j > 0 && nums[j - 1] == nums[j] && !b[j - 1])continue;
if(!b[j]){
b[j] = 1;
v[i] = nums[j];
dfs(i + 1, nums);
b[j] = 0;
}
}
return;
}
vector<vector<int>> permuteUnique(vector<int>& nums) {
sort(nums.begin(), nums.end());
v.resize(nums.size());
b.resize(nums.size());
dfs(0, nums);
return ans;
}
};
49. 字母异位词分组
思路
- stl使劲套，要用multiset，两个单词字符集相同但是字符个数不同
- 优化，不用set，map变成string，字符集的字符串排序后对应唯一的“特征字符串”
AC代码
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
map<multiset<char>, vector<string>> m;
int num = strs.size();
for (auto &x : strs) {
multiset<char> s(x.begin(), x.end());
m[s].push_back(x);
}
vector<vector<string>> ans;
for (auto &x : m) {
ans.push_back(x.second);
}
return ans;
}
};
static const auto io_sync_off = []() {
// turn off sync
std::ios::sync_with_stdio(false);
// untie in/out streams
std::cin.tie(nullptr);
return nullptr;
}();
AC代码（优化）
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string, vector<string>> m;
for (auto &x : strs) {
string temp = x;
sort(x.begin(), x.end());
m[x].push_back(temp);
}
vector<vector<string>> ans;
for (auto &x : m) {
ans.push_back(x.second);
}
return ans;
}
};
80. 删除排序数组中的重复项 II
思路
双指针
遍历一遍数组，
AC代码
class Solution {
public:
int removeDuplicates(vector<int> &nums) {
int i = 0, j = 0;
int len = nums.size();
while (i < len) {
if (j < 2 || nums[i] > nums[j - 2]) {
int n = nums[i];
nums[j++] = n;
}
i++;
}
return j;
}
};
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