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  • 剑指offer --合并链表

    题目: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。

    解法://递归解法

    public class MixLink {

    /*
    public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
    this.val = val;
    }
    }*/
    public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
    if(list1==null){
    return list2;
    }
    if(list2==null){
    return list1;
    }
    ListNode list3 = null;

    if(list1.val<list2.val){
    list3 = list1;
    list3.next = Merge(list1.next,list2);
    }else if(list1.val>=list2.val){
    list3 = list2;
    list3.next = Merge(list1,list2.next);
    }
    return list3;
    }
    }

    解法: 非递归

           while(list1!=null&&list2!=null){
                  
                 if(list1.val<list2.val){
                     if(list3==null){
                          list3 = list1;
                     }else{
                         list3.next = list1;
                         list3 = list3.next; 
                     }
                     list1 = list1.next;
                 }else if(list1.val>list2.val){
                     if(list3 == null){
                          list3 = list2;
                     }else{
                         list3.next = list2;
                         list3 = list3.next;
                     }
                     list2 = list2.next;
                 }
                  
             }
             if(list1==null){
                 list3.next= list2;
             }
             if(list2 == null){
                 list3.next = list1;
             }
             return list3;
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  • 原文地址:https://www.cnblogs.com/tandy/p/5443872.html
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