LeetCode：21. Merge Two Sorted Lists（Easy）

1. 原题链接

https://leetcode.com/problems/merge-two-sorted-lists/description/

2. 题目要求

给出两个已经从小到大排序的链表ls1、ls2，进行合并，合并后仍有序，返回合并后的链表

3. 解题思路

创建一个表头指针headPointer和一个定位指针locatePointer，headPointer用来保存头结点。

同时对ls1和ls2进行遍历，当ls1的值小于ls2的值时，locatePointer指向ls1，否则指向ls2。

当一个链表为空另一个不为空，则将不为空链表的剩余结点依次加入新的链表。

4. 代码实现

public class MergeTwoSortedList {
    public static void main(String[] args) {
        ListNode ls1 = new ListNode(-9);
        ListNode ls2 = new ListNode(3);
        ListNode ls3 = new ListNode(4);
        ListNode ls4 = new ListNode(5);
        ListNode ls5 = new ListNode(7);
        ListNode ls6 = new ListNode(8);
        ls1.next = ls2;
        ls2.next = ls3;
        ls4.next = ls5;
        ls5.next = ls6;


        ListNode ls = mergeTwoLists(ls1, ls4);
        while (ls != null) {
            System.out.println(ls.val);
            ls = ls.next;
        }
    }

    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode headPointer = new ListNode(-1);
        ListNode res = headPointer;

        while (l1 != null || l2 != null) {

            if (l1 != null && l2 == null) {
                res.next = l1;
                l1 = l1.next;
            } else if (l1 == null && l2 != null) {
                res.next = l2;
                l2 = l2.next;
            } else {
                if (l1.val < l2.val) {
                    res.next = l1;
                    l1 = l1.next;
                } else {
                    res.next = l2;
                    l2 = l2.next;
                }
            }
            res = res.next;
        }

        return headPointer.next;
    }

    public static class ListNode {
        int val;
        ListNode next;

        ListNode(int x) {
            val = x;
        }
    }
}