leetcode21：合并两个有序链表 #11
Comments
|
解答： 确定解题的数据结构： 单链表 确定解题思路： 从链表头开始比较， 画图实现： 画图帮助理解一下 确定边界条件： 当递归到任意链表为 代码实现： function mergeTwoLists(l1, l2) {
if(l1 === null) {
return l2
}
if(l2 === null) {
return l1
}
if(l1.val <= l2.val) {
l1.next = mergeTwoLists(l1.next, l2)
return l1
} else {
l2.next = mergeTwoLists(l2.next, l1)
return l2
}
} |
|
我这个就稍微麻烦了点 var mergeTwoLists = function(l1, l2) {
let res = new List();
while(l1 !== null && l2 !== null){
if(l1.element < l2.element){
res.append(l1.element);
l1 = l1.next;
}else{
res.append(l2.element);
l2 = l2.next;
}
}
let temp = !l1 ? l2 : l1;
while(temp){
res.append(temp.element);
temp = temp.next;
}
return res;
}; |
var mergeTwoLists = function(l1, l2) {
if (l1 === null) {
return l2
} else if (l2 === null) {
return l1
} else if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2)
return l1
} else {
l2.next = mergeTwoLists(l1, l2.next)
return l2
}
}; |
function ListNode(val, next) {
this.val = (val===undefined ? 0 : val)
this.next = (next===undefined ? null : next)
}
var mergeTwoLists = function(l1, l2) {
let preHead = new ListNode(-1);
let cur = preHead;
while(l1 && l2){
if(l1.val < l2.val){
cur.next = l1;
l1 = l1.next;
}else{
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = l1 || l2;
return preHead.next;
}; |
const merge = (l1, l2) => {
let temp1 = l1.head
let temp2 = l2.head
let list = new List()
while (temp1 || temp2) {
if (!temp1) {
list.add(temp2.value)
temp2 = temp2.next
} else if (!temp2) {
list.add(temp1.value)
temp1 = temp1.next
} else if (temp1.value >= temp2.value) {
list.add(temp2.value)
temp2 = temp2.next
} else {
list.add(temp1.value)
temp1 = temp1.next
}
}
return list
} |
function merge_link1(head1, head2) {
if (!head1) return head2
if (!head2) return head1
let merge_head = null
let merge_tail = null
let min_data = null
let curr1 = head1
let curr2 = head2
while (curr1 && curr2) {
if (curr1.data < curr2.data) {
min_data = curr1.data
curr1 = curr1.next
} else {
min_data = curr2.data
curr2 = curr2.next
}
if (!merge_head) {
merge_head = new Node(min_data)
merge_tail = merge_head
} else {
const new_node = new Node(min_data)
merge_tail.next = new_node
merge_tail = new_node
}
}
// 循环结束后，还有某个链表还有值
let rest_link = null
if (curr1) {
rest_link = curr1
}
if (curr2) {
rest_link = curr2
}
// while (rest_link) {
// merge_tail.next = rest_link
// merge_tail = rest_link
// rest_link = rest_link.next
// }
merge_tail.next = rest_link
return merge_head
} |
var mergeTwoLists = function(l1, l2) {
let head = new ListNode()
let cur = head
while (l1 && l2) {
if (l1.val <= l2.val) {
cur.next = l1 // 指向目标节点
l1 = l1.next // 移动到下一个节点
} else {
cur.next = l2
l2 = l2.next
}
cur = cur.next
}
// 处理l1或者l2还未遍历完的场景
cur.next = l1 || l2
return head.next
};
|
var mergeTwoLists = function (l1, l2) {
let p1 = new ListNode(0)
let end = p1
while (l1 && l2) {
if (l1.val >= l2.val) {
end.next = l2
l2 = l2.next
} else {
end.next = l1
l1 = l1.next
}
end = end.next
}
end.next = l1 ? l1 : l2
return p1.next
} |
这个图片是用什么画图工具做出来的？ |
var mergeTwoLists = function (l1, l2) {
let prev = new ListNode();
let p1 = l1, p2 = l2;
let current = prev;
while (p1 && p2) {
if (p1.val > p2.val) {
current.next = new ListNode(p2.val);
p2 = p2.next;
} else {
current.next = new ListNode(p1.val);
p1 = p1.next;
}
current = current.next;
}
current.next = p1 ? p1 :p2;
return prev.next;
}; |
题目地址(21. 合并两个有序链表)https://leetcode-cn.com/problems/merge-two-sorted-lists 题目描述前置知识公司
公司
思路本题可以使用递归来解，将两个链表头部较小的一个与剩下的元素合并，并返回排好序的链表头，当两条链表中的一条为空时终止递归。 关键点
代码
CPP Code: class Solution {
public:
ListNode* mergeTwoLists(ListNode* a, ListNode* b) {
ListNode head, *tail = &head;
while (a && b) {
if (a->val <= b->val) {
tail->next = a;
a = a->next;
} else {
tail->next = b;
b = b->next;
}
tail = tail->next;
}
tail->next = a ? a : b;
return head.next;
}
};JS Code: /**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
const mergeTwoLists = function (l1, l2) {
if (l1 === null) {
return l2;
}
if (l2 === null) {
return l1;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
};复杂度分析 M、N 是两条链表 l1、l2 的长度
扩展
大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 37K star 啦。 |
|
/**
* 重新创建一个链表
*/
var mergeTwoLists = function(l1, l2) {
if(l1===null) return l2;
if(l2===null) return l1;
let curL1 = l1;
let curL2 = l2;
let head = new ListNode(0);
let cur = head;
while(curL1!==null&&curL2!==null) {
if(curL1.val<=curL2.val) {
cur.next = curL1;
curL1 = curL1.next;
} else {
cur.next = curL2;
curL2 = curL2.next;
}
cur = cur.next;
}
cur.next = curL1===null?curL2:curL1;
return head.next;
}; |
const chain1 = {
val: 1,
next: {
val: 3,
next: {
val: 5,
next: null,
},
},
};
const chain2 = {
val: 2,
next: {
val: 4,
next: {
val: 6,
next: null,
},
},
};
function merge(l1: any, l2: any) {
if (l1 === null) {
return l2;
}
if (l2 === null) {
return l1;
}
if (l1.val < l2.val) {
l1.next = merge(l1.next, l2);
return l1;
} else {
l2.next = merge(l1, l2.next);
return l2;
}
}
merge(chain1, chain2);
console.log("chain1", chain1);
console.log("chain2", chain2); |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例：
附leetcode地址：leetcode
The text was updated successfully, but these errors were encountered: