题解 | 链表内指定区间反转

链表内指定区间反转

https://www.nowcoder.com/practice/b58434e200a648c589ca2063f1faf58c

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @param m int整型
     * @param n int整型
     * @return ListNode类
     */
    public ListNode reverseBetween (ListNode head, int m, int n) {
        // write code here
        if (m == n) {
            return head;
        }
        // 计算链表的长度
        if (lengthOfListNode(head) == 1) {
            return head;
        }
        if (lengthOfListNode(head) == 2) {
            return reverseList(head);
        }
        // 先找出要反转的链表m的前一个节点
        ListNode preM = null;
        ListNode mNode = null;
        if (preMListNode(head, m) != null) {
            preM = preMListNode(head, m);
            mNode = preM.next;
        } else {
            //说明m=1
            mNode = head;
        }
        ListNode aftN = null;
        ListNode nNode = null;
        if (nNode(head, n) != null) {
            nNode = nNode(head, n);
            aftN = nNode.next;
        }
        if (preM != null) {
            preM.next = null;
        }
        if (nNode != null) {
            nNode.next = null;
        }
        ListNode listNode = reverseList(mNode);
        if(preM!=null){
            preM.next = listNode;
        }
        mNode.next = aftN;
        if(m==1){
            return listNode;
        }
        return head;

    }

    public ListNode preMListNode (ListNode head, int m) {
        int i = 1;
        ListNode preM = null;
        while (head != null) {
            if (i == (m - 1)) {
                preM = head;
                break;
            }
            head = head.next;
            i++;
        }
        return preM;
    }

    public ListNode nNode (ListNode head, int n) {
        int i = 1;
        ListNode nNode = null;
        while (head != null) {
            if (i == n) {
                nNode = head;
                break;
            }
            head = head.next;
            i++;
        }
        return nNode;
    }

    public ListNode reverseList (ListNode head) {
        // write code here
        ListNode cur = null;
        ListNode pre = head;
        while (pre != null) {
            ListNode next = pre.next;
            pre.next = cur;
            cur = pre;
            pre = next;
        }
        return cur;
    }

    public int lengthOfListNode(ListNode head) {
        int length = 0;
        while (head != null) {
            length++;
            head = head.next;
        }
        return length;
    }
}

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利用特殊判断加链表反转
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发布于 04-12 12:27 浙江

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