LeetCode21 合并两个有序链表

• 题目
• 解题
• • 解题一：迭代
  • 解题二：递归

题目

解题

解题一：迭代

// javascript
var mergeTwoLists = function(l1, l2) {
    let mergedHead = merged = new ListNode(0);
    while (l1 !== null && l2 !== null) {
        if (l1.val < l2.val) {
            merged.next = l1;
            l1 = l1.next;
        }
        else {
            merged.next = l2;
            l2 = l2.next;
        }
        merged = merged.next;
    }
    // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
    merged.next = (l1 === null) ? l2 : l1;
    return mergedHead.next;
};

解题二：递归

// javascript
var 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;
    }
};

标签：next,链表,解题,l2,有序,l1,null,LeetCode21,merged
来源： https://blog.csdn.net/weixin_45561634/article/details/120629004