链表

难度简单

题目描述

解题思路

利用两个指针，pre指向前一个节点，cur指向当前节点，就地反转

Code

class ListNode {
    int val;
    ListNode next;
    ListNode() {
    }
    ListNode(int val) {
        this.val = val;
    }
    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}
public ListNode reverseList(ListNode head) {
    ListNode pre = null, cur = head;
    while (cur != null) {
        ListNode temp = cur.next;
        cur.next = pre;
        pre = cur;
        cur = temp;
    }
    return pre;
}