题解 | #合并两个排序的链表#

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode Merge(ListNode list1, ListNode list2) {
	  //设置返回链表
        ListNode head = new ListNode(Integer.MIN_VALUE);
	  //当链表完成合并后head不再指向原地址，所以用curr存储原地址
        ListNode curr = head;
	  //遍历循环大小，排序
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                head.next = list1;
                list1 = list1.next;
            } else {
                head.next = list2;
                list2 = list2.next;
            }
            head = head.next;
        }
        if (list1 != null)
            head.next = list1;
        if (list2 != null)
            head.next = list2;
	  //从链表要返回原地址的next，才是真正的排序好的新链表
        return curr.next;
    }
}