2022.07.12 _ a day

1669. 合并两个链表

题目描述

给你两个链表 list1 和 list2 ,它们包含的元素分别为 n 个和 m 个.

请你将 list1 中下标从 a 到 b 的全部节点都删除,并将list2 接在被删除节点的位置.

下图中蓝色边和节点展示了操作后的结果：

请你返回结果链表的头指针.

 

示例 1：

输入：list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
输出：[0,1,2,1000000,1000001,1000002,5]
解释：我们删除 list1 中下标为 3 和 4 的两个节点,并将 list2 接在该位置.上图中蓝色的边和节点为答案链表.

示例 2：

输入：list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
输出：[0,1,1000000,1000001,1000002,1000003,1000004,6]
解释：上图中蓝色的边和节点为答案链表.

 

提示：

• 3 <= list1.length <= 104
• 1 <= a <= b < list1.length - 1
• 1 <= list2.length <= 104

Related Topics

• 链表

coding

//leetcode submit region begin(Prohibit modification and deletion)
/** * 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 mergeInBetween1(ListNode list1, int a, int b, ListNode list2) {
    
		ListNode res = list1;
		for (int i = 1; i < a; i++) {
    
			list1 = list1.next;
		}
		// remove <=> 左边界
		ListNode remove = list1;
		for (int i = 0; i < b - a + 1; i++) {
    
			remove = remove.next;
		}
		// remove <=> 右边界

		// 左边界 -> list2
		while (list2 != null) {
    
			list1.next = list2;
			list1 = list1.next;
			list2 = list2.next;
		}
		// list2 结束 -> list1 Right part after element is removed
		list1.next = remove.next;
		return res;
	}
		
	public ListNode mergeInBetween2(ListNode list1, int a, int b, ListNode list2) {
    
		ListNode cur = list1;
		ListNode left = list1;
		ListNode right = list1;
		// 记录左右边界
		for (int i = 0; cur != null; i++) {
    
			if (a - 1 == i) {
    
				left = cur;
			}
			if (b + 1 == i) {
    
				right = cur;
			}
			cur = cur.next;
		}
		left.next = list2;
		while (list2.next != null) {
    
			list2 = list2.next;
		}
		list2.next = right;
		return list1;
	}
}
//leetcode submit region end(Prohibit modification and deletion)