1、单链表翻转
/**
* Definition of singly-linked-list:
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: n
* @return: The new head of reversed linked list.
*/
ListNode* reverse(ListNode *head) {
// write your code here
if (head == nullptr || head->next == nullptr)
return head;
ListNode* pPre = head;
ListNode* pCur = head->next;
head->next = nullptr;
while (pCur) {
ListNode* pTemp = pCur->next;
pCur->next = pPre;
pPre = pCur;
pCur = pTemp;
}
return pPre;
}
};
核心要点：一定要注意将首节点的next指针赋值为nullptr，否则会导致死循环。
题目链接：35 · 翻转链表（一） - LintCode
1、链表翻转2
/**
* Definition of singly-linked-list:
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: ListNode head is the head of the linked list
* @param m: An integer
* @param n: An integer
* @return: The head of the reversed ListNode
*/
ListNode* reverseBetween(ListNode *head, int m, int n) {
// write your code here
if (!head || !head->next || m == n)
return head;
ListNode tempHead(0);
tempHead.next = head;
ListNode* pPre = &tempHead;
ListNode* pCur = head;
ListNode* pMPre = nullptr;
int count = 1;
while (pCur) {
if (count == m) {
pMPre = pPre;
break;
}
pPre = pCur;
pCur = pCur->next;
++count;
}
pMPre->next = nullptr;
ListNode* pNPre = pCur;
while (pCur) {
if (count == n) {
pNPre->next = pCur->next;
}
ListNode* pNext = pCur->next;
pCur->next = pMPre->next;
pMPre->next = pCur;
pCur = pNext;
if (count == n)
break;
++count;
}
return tempHead.next;
}
};
核心要点：
1）创建一个哨兵节点指向头结点，可以方便处理 m==1的情况，哨兵节点可以使用栈对象，使用new还需要delete；
2）两次while循环，第一次找到m节点的上一个节点pMPre，此时pCur节点是n节点的上一个节点，需要记录此节点的指针ListNode* pNPre = pCur，还要注意pMPre->next = nullptr;
- 第二次while循环翻转链表，并查找n节点，找到n节点停止，并pNPre->next = pCur->next;
- 注意此时头结点的指针是tempHead.next，而不是head;
- 当只有一个节点，或m==n时不做任何操作，直接返回head
3、回文链表
/**
* Definition of singly-linked-list:
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: A ListNode.
* @return: A boolean.
*/
bool isPalindrome(ListNode *head) {
// write your code here
if (!head || !head->next)
return true;
ListNode* pSlow = head;
ListNode* pFast = head;
while (pFast->next && pFast->next->next) {
pFast = pFast->next->next;
pSlow = pSlow->next;
}
ListNode* pCur = pSlow->next;
ListNode tempHead(0);
tempHead.next = nullptr;
while (pCur) {
ListNode* pNext = pCur->next;
pCur->next = tempHead.next;
tempHead.next = pCur;
pCur = pNext;
}
pSlow->next = nullptr;
ListNode* pLeft = head;
ListNode* pRight = tempHead.next;
while (pLeft && pRight) {
if (pLeft->val != pRight->val)
return false;
pLeft = pLeft->next;
pRight = pRight->next;
}
return true;
}
};
核心要点：一定要翻转后半部分，如果翻转前半部分，需要考虑节点个数是奇数还是偶数，逻辑会非常复杂，翻转后半部分不需要考虑这些，因为不管节点个数是奇数还是偶数，后半部分的第一个节点都是pSlow的下一个节点。
题目链接：223 · 回文链表 - LintCode
4、第k大元素
class Solution {
public:
/**
* @param k: An integer
* @param nums: An array
* @return: the Kth largest element
*/
int kthLargestElement(int k, vector<int> &nums) {
// write your code here
int begin = 0;
int end = nums.size() - 1;
return quicksort(nums, begin, end, k);
}
int quicksort(vector<int> &nums, int begin, int end, int k) {
if (begin == end)
return nums[begin];
int partion = nums[end];
int i = begin;
for (int j = begin; j < end; j++) {
if (nums[j] >= partion) {
int temp = nums[j];
nums[j] = nums[i];
nums[i] = temp;
++i;
}
}
nums[end] = nums[i];
nums[i] = partion;
if (k == i + 1)
return nums[i];
else if (i < k)
return quicksort(nums, i + 1, end, k);
else
return quicksort(nums, begin, i - 1, k);
}
};
核心要点：i之前的元素都是大于分割点的元素，j遍历的终点是end-1，最后第i个元素要和end元素交换。
题目链接：5 · 第k大元素 - LintCode
5、在排序数组中查找元素的第一个和最后一个位置
class Solution {
public:
/**
* @param nums: the array of integers
* @param target:
* @return: the starting and ending position
*/
vector<int> searchRange(vector<int> &nums, int target) {
// Write your code here.
vector<int> result;
int low = 0;
int high = nums.size() - 1;
int tmp_low = 0;
int tmp_high = high;
bool first = true;
while (low <= high) {
int mid = low + ((high - low) >> 1);
if (nums[mid] > target) {
high = mid - 1;
} else if (nums[mid] < target) {
low = mid + 1;
} else {
if (first) {
tmp_low = low;
tmp_high = high;
first = false;
}
if (mid == 0 || nums[mid - 1] < target) {
result.push_back(mid);
break;
}
high = mid - 1;
}
}
if (result.empty()) {
result.push_back(-1);
result.push_back(-1);
return result;
}
low = tmp_low;
high = tmp_high;
while (low <= high) {
int mid = low + ((high - low) >> 1);
if (nums[mid] > target) {
high = mid - 1;
} else {
if (mid == nums.size() - 1 || nums[mid + 1] > target) {
result.push_back(mid);
break;
}
low = mid + 1;
}
}
return result;
}
};
核心要点：注意运算符的优先级 int mid = low + ((high - low) >> 1); 而不是 int mid = low + (high - low) >> 1; 否则会死循环。
题目链接：1536 · 在排序数组中查找元素的第一个和最后一个位置 - LintCode
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