【Leetcode problrm 21】Merge Two Sorted Lists
2017-08-17 15:42
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思路:将l1、l2两个列表合并到新列表中
1.建立一个llistnode对象list用于表示合并后的列表,再建一个temp用于保存列表中的当前值。
2.判断l1和l2当前值的大小,然后将list。next指向较小的值。
3.循环计算。
javascript代码如下:
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function(l1, l2) {
var list=new ListNode(0);
var temp=list;
while(l1!==null&&l2!==null){
if(l1.val>l2.val){
temp.next=l2;
l2=l2.next;
temp=temp.next;
}else{
temp.next=l1;
l1=l1.next;
temp=temp.next;
}
}
if(l1!==null){
temp.next=l1;
}
if(l2!==null){
temp.next=l2;
}
return list.next;
};
1.建立一个llistnode对象list用于表示合并后的列表,再建一个temp用于保存列表中的当前值。
2.判断l1和l2当前值的大小,然后将list。next指向较小的值。
3.循环计算。
javascript代码如下:
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function(l1, l2) {
var list=new ListNode(0);
var temp=list;
while(l1!==null&&l2!==null){
if(l1.val>l2.val){
temp.next=l2;
l2=l2.next;
temp=temp.next;
}else{
temp.next=l1;
l1=l1.next;
temp=temp.next;
}
}
if(l1!==null){
temp.next=l1;
}
if(l2!==null){
temp.next=l2;
}
return list.next;
};
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