题解 | #反转链表#
反转链表
https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* ReverseList(ListNode* pHead) {
while(!pHead || !pHead->next) return pHead;
ListNode *pre, *cur, *nex;
pre = NULL;
cur = pHead;
while(cur) {
nex = cur->next;
cur->next = pre;
pre = cur;
cur = nex;
}
return pre;
}
};
三指针最简单思路，pre最开始指向NULL，cur为当前需要修改链接的节点，nex记录下一个待遍历节点
顺序为：
nex先记录cur的后继节点，
断开指针指向前节点pre，
pre前移指向cur，cur前移指向nex
