题解 | #合并两个排序的链表#

有链表为空时，直接连接不为空的链表；都不为空时比较大小，跟新指针

/*
public class ListNode {
    int val;
    ListNode next = null;
    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode Merge(ListNode list1, ListNode list2) {
        ListNode res = new ListNode(-1);
        ListNode resNext = res.next;
        int j = 0;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                res.next = list1;
                list1 = list1.next;
                res = res.next;
                j++;
            } else {
                res.next = list2;
                list2 = list2.next;
                res = res.next;
                j++;
            }
            if (j == 1)
                resNext = res;
        }
        if (list1 == null && list2 != null) {
            res.next = list2;
            res = res.next;
            j++;
            if (j == 1)
                resNext = res;
        }
        if ( list2 == null && list1 != null ) {
            res.next = list1;
            res = res.next;
            j++;
            if (j == 1)
                resNext = res;
        }
        return resNext;
    }
}