合并两个有序单链表

我会带着你远行 2022-11-30 04:18 198阅读 0赞

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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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