原文链接: https://leetcode-cn.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target

英文原文

Given an array of digit strings nums and a digit string target, return the number of pairs of indices (i, j) (where i != j) such that the concatenation of nums[i] + nums[j] equals target.

 

Example 1:

Input: nums = ["777","7","77","77"], target = "7777"
Output: 4
Explanation: Valid pairs are:
- (0, 1): "777" + "7"
- (1, 0): "7" + "777"
- (2, 3): "77" + "77"
- (3, 2): "77" + "77"

Example 2:

Input: nums = ["123","4","12","34"], target = "1234"
Output: 2
Explanation: Valid pairs are:
- (0, 1): "123" + "4"
- (2, 3): "12" + "34"

Example 3:

Input: nums = ["1","1","1"], target = "11"
Output: 6
Explanation: Valid pairs are:
- (0, 1): "1" + "1"
- (1, 0): "1" + "1"
- (0, 2): "1" + "1"
- (2, 0): "1" + "1"
- (1, 2): "1" + "1"
- (2, 1): "1" + "1"

 

Constraints:

• 2 <= nums.length <= 100
• 1 <= nums[i].length <= 100
• 2 <= target.length <= 100
• nums[i] and target consist of digits.
• nums[i] and target do not have leading zeros.

中文题目

给你一个 数字 字符串数组 nums 和一个 数字 字符串 target ，请你返回 nums[i] + nums[j] （两个字符串连接）结果等于 target 的下标 (i, j) （需满足 i != j）的数目。

 

示例 1：

输入：nums = ["777","7","77","77"], target = "7777"
输出：4
解释：符合要求的下标对包括：
- (0, 1)："777" + "7"
- (1, 0)："7" + "777"
- (2, 3)："77" + "77"
- (3, 2)："77" + "77"

示例 2：

输入：nums = ["123","4","12","34"], target = "1234"
输出：2
解释：符合要求的下标对包括
- (0, 1)："123" + "4"
- (2, 3)："12" + "34"

示例 3：

输入：nums = ["1","1","1"], target = "11"
输出：6
解释：符合要求的下标对包括
- (0, 1)："1" + "1"
- (1, 0)："1" + "1"
- (0, 2)："1" + "1"
- (2, 0)："1" + "1"
- (1, 2)："1" + "1"
- (2, 1)："1" + "1"

 

提示：

• 2 <= nums.length <= 100
• 1 <= nums[i].length <= 100
• 2 <= target.length <= 100
• nums[i] 和 target 只包含数字。
• nums[i] 和 target 不含有任何前导 0 。

通过代码

高赞题解

就按照题目描述来写

1. i != j
2. nums [ i ] + nums [ j ] = target ( 必须满足条件一 )

然后我就想两个for循环暴力过去，代码如下：

class Solution:
    def numOfPairs(self, nums: List[str], target: str) -> int:
        n = len ( nums )
        res = 0
        for i in range ( n ) :
            for j in range ( n ) :
                if i != j :
                    if nums [ i ] + nums [ j ] == target :
                        res += 1
        return res

跑起来的结果是我没有想到的(可以算是混过去了？)

题解要是有哪些地方不对的，各位大佬可以提出来；若是有不同解法，也欢迎提出来

题外话 ：我要被自己蠢哭了，呜呜呜，Markdown语法好难

统计信息

通过次数	提交次数	AC比率
3122	4167	74.9%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言