剑指Offer

    xiaoxiao2025-03-06  6

    题目描述

    输入两个单调递增的链表,输出两个链表合成后的链表,当然我们需要合成后的链表满足单调不减规则。

    解题思路

    依次从头开始遍历,比较大小,将小的结点插入新的链表,并将指针后移,直到两个链表遍历完成。

    实现

    /*链表结点定义*/ public class ListNode { int val; ListNode next = null; ListNode(int val) { this.val = val; } } /*实现*/ public class Solution { public ListNode Merge(ListNode list1,ListNode list2) { ListNode head = null, p = head; ListNode p1 = list1, p2 = list2; while (p1 != null && p2 != null){ if (p1.val <= p2.val){ if (head == null){ head = p = p1; }else { p.next = p1; p = p.next; } p1 = p1.next; }else{ if (head == null){ head = p = p2; }else { p.next = p2; p = p.next; } p2 = p2.next; } } if (p1 != null){ if (head == null){ head = p = p1; }else { p.next = p1; } } if (p2 != null){ if (head == null){ head = p = p2; }else { p.next = p2; } } return head; } }
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