day36（12.17）——leetcode面试经典150

21. 合并两个有序链表

21. 合并两个有序链表

我的微服务还没正式开启呢，我这两天在做一个贼简单的网站，昨天写的很兴奋，写到三点，整的今天巨困！！！

题目：

题解：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode cur = new ListNode(0);
        ListNode list = cur;
        while(list1!=null && list2!=null) {
            if(list1.val<list2.val) {
                cur.next = new ListNode(list1.val);
                list1 = list1.next; 
            }
            else {
                cur.next = new ListNode(list2.val);
                list2 = list2.next;
            }
            cur = cur.next;
        }
        while(list1 != null) {
            cur.next = new ListNode(list1.val);
            list1 = list1.next;
            cur = cur.next;
        }
        while(list2 != null) {
            cur.next = new ListNode(list2.val);
            list2 = list2.next;
            cur = cur.next;
        }
        return list.next;
    }
}

还有另一种写法，思路差不多

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode cur = new ListNode(0);
        ListNode list = cur;
        while(list1!=null || list2!=null) {
            if(list1 == null && list2 != null) {
                cur.next = new ListNode(list2.val);
                list2 = list2.next;
            }
            else if(list1 != null && list2 == null) {
                cur.next = new ListNode(list1.val);
                list1 = list1.next;
            }
            //只剩下两个都存在的情况了
            //两个都不存在就进来循环了
            else {
                if(list1.val<list2.val) {
                    cur.next = new ListNode(list1.val);
                    list1 = list1.next; 
                }
                else {
                    cur.next = new ListNode(list2.val);
                    list2 = list2.next;
                }
            }
            cur = cur.next;
        }
        return list.next;
    }
}