标签：font put lan struct out for div turn lis
https://leetcode.com/problems/reverse-linked-list-ii/
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL
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) {
if(!head) return NULL;
int num = m - 1;
ListNode *dummy = new ListNode(-1);
ListNode *st = dummy, *ans = st;;
dummy -> next = head;
ListNode *slow = head;
n -= m, m -= 1;
while(m --) slow = slow -> next;
ListNode *cur = slow;
while(n --) cur = cur -> next;
ListNode *end = cur -> next;
cur -> next = NULL;
slow = reverseList(slow);
ListNode *fast = slow;
while(fast -> next) fast = fast -> next;
fast -> next = end;
while(num --) st = st -> next;
st -> next = slow;
return ans -> next;
}
ListNode *reverseList(ListNode *c) {
if(!c || !c -> next) return c;
ListNode *New = reverseList(c -> next);
c -> next -> next = c;
c -> next = NULL;
return New;
}
};
分三段 然后拼在一起 但是写的乱糟糟 大概写了一个多小时 我恨！自己不争气 落泪
FH
#Leetcode# 92. Reverse Linked List II
标签：font put lan struct out for div turn lis
原文地址：https://www.cnblogs.com/zlrrrr/p/10409250.html
