高频面试题总结之翻转链表

数据结构
数据结构, 高频面试题总结

高频面试题总结 —— 翻转链表

翻转链表是面试中常见的题目，这并不是什么难题，因为要实现翻转，所以一般需要有两个ListNode，通过改变这两个ListNode的前后节点及循环移动这两个节点从而实现链表的翻转。

初级翻转链表

给出一个链表1->2->3->null，这个翻转后的链表为3->2->1->null

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/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */
public class Solution {
    /*
     * @param head: n
     * @return: The new head of reversed linked list.
     */
     
    public ListNode reverse(ListNode head) {
        // write your code here
        /**
         * 1.
         * 
         * */
        /**非递归实现
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode cur = head;
        ListNode dummy = new ListNode(-1);
        
        while (cur != null) {
            ListNode temp = cur.next;
            cur.next = dummy.next;
            dummy.next = cur;
            cur = temp;
        }
        
        return dummy.next;**/
        
        //递归实现
        if (head == null || head.next == null) {
            return head;
        }
        ListNode nextNode = head.next;
        head.next = null;
        ListNode reverseResourse = reverse(nextNode);
        nextNode.next = head;
        return reverseResourse;
    }
}

翻转部分链表

翻转链表中第m个节点到第n个节点的部分
给出链表1->2->3->4->5->null， m = 2 和n = 4，返回1->4->3->2->5->null

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/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param head: ListNode head is the head of the linked list 
     * @param m: An integer
     * @param n: An integer
     * @return: The head of the reversed ListNode
     */
    public ListNode reverseBetween(ListNode head, int m, int n) {
        // write your code here
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode cur = head;
        for (int i = 1; i < m; i++) {
            pre = pre.next;
            cur = cur.next;
        }
        //与翻转链表的不同是要保持第一个节点不变
        for (int i = 0; i < n - m; i++) {
            ListNode temp = cur.next;
            cur.next = temp.next;
            temp.next = pre.next;
            pre.next = temp;
        }
        return dummy.next;
    }
}
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