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21.合并两个有序链表 (opens new window)

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Langs: c cpp csharp golang java javascript kotlin php python python3 ruby rust scala swift typescript

• algorithms
• Easy (64.95%)
• Likes: 1446
• Dislikes: -
• Total Accepted: 433.9K
• Total Submissions: 668.1K
• Testcase Example: '[1,2,4]\n[1,3,4]'

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：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

/*
 * @lc app=leetcode.cn id=21 lang=javascript
 *
 * [21] 合并两个有序链表
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function (l1, l2) {
    const hair = new ListNode(0, l1);
    let prev = hair, p;
    while (l1 && l2) {
        if (l2.val < l1.val) {
            p = l2.next;
            l2.next = prev.next;
            prev.next = l2;
            l1 = l2;
            l2 = p;
        } else {
            prev = l1;
            l1 = l1.next;
        }
    }
    l2 && (prev.next = l2);

    return hair.next;
};
// 递归
var mergeTwoLists2 = function (l1, l2) {
    if(l1 === null){
        return l2;
    }else if(l2 === null){
        return l1;
    }else if(l1.val <= l2.val){
        l1.next = mergeTwoLists2(l1.next,l2);
        return l1;
    }else{
        l2.next = mergeTwoLists2(l1,l2.next);
        return l2;
    }
};
// @lc code=end


// @after-stub-for-debug-begin
module.exports = mergeTwoLists;
// @after-stub-for-debug-end

/*
 * @lc app=leetcode.cn id=21 lang=javascript
 *
 * [21] 合并两个有序链表
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var mergeTwoLists = function(l1, l2) {
    const dummy = new ListNode()
    let p = dummy
    while (l1 && l2) {
        if (l1.val < l2.val) {
            p.next = l1
            l1 = l1.next
            p = p.next
        } else {
            p.next = l2
            l2 = l2.next
            p = p.next
        }
    }

    if (l1) {
        p.next = l1
    } else if (l2) {
        p.next = l2
    }

    return dummy.next
}

// @lc code=end

// @after-stub-for-debug-begin
module.exports = mergeTwoLists
// @after-stub-for-debug-end

/*
 * @lc app=leetcode.cn id=21 lang=javascript
 *
 * [21] 合并两个有序链表
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var mergeTwoLists = function(l1, l2) {
    if (!l1 && !l2) {
        return null
    } else if (!l1) {
        return l2
    } else if (!l2) {
        return l1
    }

    const dummy = new ListNode()
    let p = dummy
    if (l1.val < l2.val) {
        p.next = l1
        l1 = l1.next
    } else {
        p.next = l2
        l2 = l2.next
    }

    p.next.next = mergeTwoLists(l1, l2)
    return dummy.next
}

// @lc code=end

// @after-stub-for-debug-begin
module.exports = mergeTwoLists
// @after-stub-for-debug-end