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一、问题描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

二、代码实现

/*public class ListNode {    int val;    ListNode next = null;    ListNode(int val) {        this.val = val;    }}*/public class Solution {        public ListNode Merge(ListNode list1,ListNode list2) {        if (list1 == null) {            return list2;        }        if (list2 == null) {            return list1;        }                ListNode newHead = (list1.val <= list2.val) ? list1 : list2;        if (list1.val <= list2.val) {            newHead = list1;            list1 = list1.next;    //无需加上newHead.next = null; 因为此场景不可能出现环        } else {            newHead = list2;            list2 = list2.next;        }        ListNode newTail = newHead;                while (list1 != null && list2 != null) {            if (list1.val <= list2.val) {                newTail.next = list1;                list1 = list1.next;            } else {                newTail.next = list2;                list2 = list2.next;            }            newTail = newTail.next;        }        if (list1 != null) {            newTail.next = list1;        } else {            newTail.next = list2;        }        return newHead;    }        public ListNode Merge1(ListNode list1,ListNode list2) {        if (list1 == null) {            return list2;        }        if (list2 == null) {            return list1;        }                ListNode newHead = null, newTail = null;        while (list1 != null && list2 != null) {            if (list1.val <= list2.val) {                if (newHead == null) {                    newHead = list1;                    newTail = list1;                } else {                    newTail.next = list1;                    newTail = list1;                }                list1 = list1.next;            } else {                if (newHead == null) {                    newHead = list2;                    newTail = list2;                } else {                    newTail.next = list2;                    newTail = list2;                }                list2 = list2.next;            }        }        if (list1 != null) {            newTail.next = list1;        }        if (list2 != null) {            newTail.next = list2;        }        return newHead;    }}

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