专栏首页搬砖记录57 Find the Distance Value Between Two Arrays

57 Find the Distance Value Between Two Arrays

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

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