题目

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays.

The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d.

Example 1:

Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
Output: 2
Explanation:
For arr1[0]=4 we have:
|4-10|=6 > d=2
|4-9|=5 > d=2
|4-1|=3 > d=2
|4-8|=4 > d=2
For arr1[1]=5 we have:
|5-10|=5 > d=2
|5-9|=4 > d=2
|5-1|=4 > d=2
|5-8|=3 > d=2
For arr1[2]=8 we have:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2

Example 2:

Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
Output: 2

Example 3:

Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
Output: 1

Constraints:

1 <= arr1.length, arr2.length <= 500
-10^3 <= arr1[i], arr2[j] <= 10^3
0 <= d <= 100

分析

题意：给定一个arr1数组，一个arr2数组，以及距离d；
找出使 |arr1[i]-arr2[j]| <= d 成立的 i。

先根据题意直接暴力遍历。

class Solution {
    public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
        for(int i=0;i<arr1.length;++i){
            for(int j=0;j<arr2.length;++j){
                if(Math.abs(arr1[i]-arr2[j])<=d)
                    return i;
            }
        }
        return -1;
    }
}

他这个题意可能描述的不准确，我跑了之后答案错误，查看了别人的答案，思路相同，但是返回值处理不同。

class Solution {
    public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
        int res = 0;
        for (int i = 0; i < arr1.length; i++) {
            boolean pass = true;
            for (int j = 0; j < arr2.length; j++) {
                if (Math.abs(arr1[i] - arr2[j]) <= d) {
                    pass = false;
                    break;
                }
            }
            if (pass) res++;
        }
        return res;
    }
}

他这个是返回通过|arr[i]-arr[j]|>d 验证的arr[i]个数。
对于例一，[4,5,8],[4,5]通过了，所以答案是2
对于例二，[1,4,2,3],[1,2]通过了，所以答案是2

这个是我英语水平不行。。

解答

如上