LeetCode.21_[链表]_合并两个有序链表

1.题目

2.解题思路

我的解法：将两个列表第一个元素较小的作为主链，之后把副链的元素插入到主链
其他解法：设置一条新的链表，之后将两个链表元素比较，较小的插入新链表链尾，插入之后对应链表移动到下一节点

3.代码实现

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode temp1 = l1;
        ListNode temp2 = l2;
        ListNode cur = l1;
        if(l1 != null && l2 != null){
            if(l1.val > l2.val){
                temp1 = l2;
                temp2 = l1;
                cur = l2;
            }
        }else if(l1 == null){
            cur = l2;
        }else{
            cur = l1;
        }
      
        while(temp1 != null && temp2 != null) {
			int num = temp2.val;
			if(num >= temp1.val) {
				ListNode temp = new ListNode(num);
				if(temp1.next == null) {
					temp1.next = temp2;
					break;
				}else if(num <= temp1.next.val){
					temp.next = temp1.next;
					temp1.next = temp;
					temp2 = temp2.next;
				}
			}
			temp1 = temp1.next;
		}
        return cur;
    }
}

4.其他解法

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        // 类似归并排序中的合并过程
        ListNode dummyHead = new ListNode(0);
        ListNode cur = dummyHead;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                cur = cur.next;
                l1 = l1.next;
            } else {
                cur.next = l2;
                cur = cur.next;
                l2 = l2.next;
            }
        }
        // 任一为空，直接连接另一条链表
        if (l1 == null) {
            cur.next = l2;
        } else {
            cur.next = l1;
        }
        return dummyHead.next;
    }
    //来源于leetcode用户Angus-Liu