每日一题:
链表
学习内容:
具体code:
分析:
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
class Solution {
public:
/**
*
* @param head ListNode类
* @param m int整型
* @param n int整型
* @return ListNode类
*/
ListNode* reverseBetween(ListNode* head, int m, int n) {
// write code here
ListNode* res = new ListNode(0);
res->next = head;
ListNode* pre = res;
ListNode* cur = head;
for (int i = 1; i < m; i++) {
pre = cur;
cur = cur->next;
}
for (int i = m; i < n; i++) {
ListNode* tmp = cur->next;
cur->next = tmp->next;
tmp->next = pre->next;
pre->next = tmp;
}
return res->next;
}
};