leetcode LinkList专题

来源：互联网发布：数据库建表工具编辑：程序博客网时间：2024/05/22 13:51

此次blog会将leetcode上的linklist专题内容放在这里，后续慢慢添加

一：leetcode 206 Reverse Linked List 二：leetcode 92 Reverse Linked List II

一：leetcode 206 Reverse Linked List

题目：

Reverse a singly linked list.

代码：

class Solution {public:    ListNode* reverseList(ListNode* head) {        if(head == NULL) return NULL;        ListNode *p = head;        ListNode *pNext = p->next;        p->next = NULL;       // 头结点需要指向NULL 否则time limit        while(pNext != NULL){            ListNode *q = pNext->next;            pNext->next = p;            p = pNext;   // 迭代            pNext = q;        }        return p;            }};

二：leetcode 92 Reverse Linked List II

题目：

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

分析：此题比一稍微复杂一点，但是也是很容易求解的

代码：

class Solution {public:    ListNode* reverseBetween(ListNode* head, int m, int n) {        if(m == n) return head;        ListNode *pre = head;        ListNode *p = head;        for(int i = 1; i < m; i++){    // 找到第m个元素和其前一个元素            pre = p;            p = p->next;        }        ListNode *curr = p;        ListNode *pNext = p->next;   // 对中间需要进行reverse的元素进行反转 利用reverse linked list中的思想        for(int i = m; i < n; i++){            ListNode *q = pNext->next;            pNext->next = p;            p = pNext;            pNext = q;        }        curr->next = pNext;        if(pre != curr)pre->next = p;      // 不相等表明是m!=1  那么 则pre->next = p,,相等表明m==1，那么head ==p        else head = p;        return head;            }};

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