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Copy path21.merge-two-sorted-lists.java
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80 lines (70 loc) · 2.2 KB
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/*
* @lc app=leetcode id=21 lang=java
*
* [21] Merge Two Sorted Lists
*/
// @lc code=start
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummyNode = new ListNode(-1);
ListNode currNode = dummyNode;
while (l1 != null && l2 != null) {
// 如果l1的值大于l2,则把l2放到当前node后面
if (l1.val > l2.val) {
currNode.next = l2;
l2 = l2.next;
}
// 否则把l1放到当前node后面
else {
currNode.next = l1;
l1 = l1.next;
}
currNode = currNode.next;
}
if (l1 != null) {
currNode.next = l1;
}
if (l2 != null) {
currNode.next = l2;
}
return dummyNode.next;
}
// 直接在l1上操作,转了个弯面试时这样写容易犯错
// public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
// // 默认主链是l1,当前node就是在遍历l1
// ListNode dummyNode = new ListNode(-1, l1);
// ListNode currNode = dummyNode;
// while (l1 != null && l2 != null) {
// // 如果l1的值大于l2,则把l2插入l1前面
// if (l1.val > l2.val) {
// // 把l2放到当前node的后面
// // 并把原本当前node的后面接到l2的后面
// ListNode nextNode = l2.next;
// l2.next = currNode.next;
// currNode.next = l2;
// l2 = nextNode;
// }
// // 否则不需要对当前l1的node做任何修改
// // l1移到下一个node
// else {
// l1 = l1.next;
// }
// currNode = currNode.next;
// }
// if (l2 != null) {
// currNode.next = l2;
// }
// return dummyNode.next;
// }
}
// @lc code=end