92.反转链表||

1. 题目描述2.代码如下

原题目连接

1. 题目描述

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。 说明: 1 ≤ m ≤ n ≤ 链表长度。 示例: 输入: 1->2->3->4->5->NULL, m = 2, n = 4 输出: 1->4->3->2->5->NULL

2.代码如下

//2019-04-22 typedef struct ListNode node; struct ListNode* reverseBetween(struct ListNode* head, int m, int n) { if((m == n )||(head->next == NULL)||(head == NULL)) { return head; } node *ptemp = head; node *ppre = NULL; node *pafter = NULL; node *pm = NULL; node *pn = NULL; int i = 1; pn = pm = head; while(i < m) { ppre = ptemp; ptemp = ptemp->next; pm = ptemp; i++; } while(i < n) { ptemp = ptemp->next; pn = ptemp; pafter = pn->next; i++; } node *temp = NULL; node *pmflag = pm; node *pnflag = pn; node *pnewhead = (node*)malloc(sizeof(node)); memset(pnewhead,0,sizeof(node)); if(pm&&pn) { ptemp = pm; temp = ptemp->next; while(ptemp != pafter) { ptemp->next = pnewhead; pnewhead = ptemp; ptemp = temp; if(temp!=NULL) { temp = temp->next; } } pmflag->next = pafter; if(ppre != NULL) {//m!=1 ppre->next = pnewhead; } else {//m == 1 head = pnflag; } } return head; } ```c