剑指offer 66道：合并两个排序的链表

合并两个排序的链表

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。

思路

分别用python和javascript实现
递归来实现，考虑两个链表是空的时

github

python代码链接: https://github.com/seattlegirl/jianzhioffer/blob/master/merge-two-sorted-linked.py.

题目代码（python）

# -*- coding:utf-8 -*-
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        if pHead1==None:
            return pHead2
        elif pHead2==None:
            return pHead1
        mergeHead=None
        if pHead1.val<pHead2.val:
            mergeHead=pHead1
            mergeHead.next=self.Merge(pHead1.next,pHead2)
        else:
            mergeHead=pHead2
            mergeHead.next=self.Merge(pHead1,pHead2.next)
        return mergeHead

x=ListNode(1)
x.next=ListNode(3)
x.next.next=ListNode(5)
x.next.next.next=ListNode(7)

y=ListNode(2)
y.next=ListNode(4)
y.next.next=ListNode(6)
y.next.next.next=ListNode(10)

res=Solution().Merge(x,y)
while res:
    print(res.val)
    res=res.next

github

JavaScript代码链接: https://github.com/seattlegirl/jianzhioffer/blob/master/merge-two-sorted-linked.js.

题目代码（JavaScript）

function ListNode(x){
    this.val = x;
    this.next = null;
}
function FindKthToTail(head, k)
{
    // 链表为空时，k为0时
    if(head===null || k==0){
        return 0;
    }
    var first=head;
    var second;
    for(var i=0;i<k-1;i++){
        if(first.next!=null){//判断节点个数是否小于k
            first=first.next;
        }
        else{
            return 0;
        }
    }
    second=head;
    while(first.next!=null){
        first=first.next;
        second=second.next;
    }
    return second;
function ListNode(x){
    this.val = x;
    this.next = null;
}
function Merge(pHead1, pHead2)
{
    if(pHead1===null){
        return pHead2;
    }
    else if(pHead2===null){
        return pHead1;
    }
    var mergeHead=null;
    if(pHead1.val<pHead2.val){
        mergeHead=pHead1;
        mergeHead.next=Merge(pHead1.next,pHead2);
    }
    else{
        mergeHead=pHead2;
        mergeHead.next=Merge(pHead1,pHead2.next);
    }
    return mergeHead;
}

var x1=new ListNode(1);
var x2=new ListNode(3);
var x3=new ListNode(5);
x2.next=x1.next;
x1.next=x2;
x3.next=x2.next;
x2.next=x3;


var y1=new ListNode(2);
var y2=new ListNode(4);
var y3=new ListNode(6);
y2.next=y1.next;
y1.next=y2;
y3.next=y2.next;
y2.next=y3;

var res=Merge(x1,y1);
// console.log(res.val);
while(res){
    console.log(res.val);
    res=res.next;
}