Leetcode- > 2-point search and clock in (3)

Effective complete square number

 public boolean isPerfectSquare(int num) {
        int left = 1;
        int right = num;
        while(left<=right){
            int middle = left+(right-left)/2;
            long tem = (long)middle*middle;
            if(tem==num){
                return true;
            }else if(tem>num){
                    right = middle-1;
            }else{
                    left = middle+1;
            }
        }
        return false;
    }

The distance between two arrays

The key to this problem is -> Using dichotomy in arr2 Find the distance arr1[X] The closest value , But the value to be found is the first greater than or equal to arr1[x], also , The first number before the value is less than or equal to arr1[x] Value .

At the same time, if we are arr2 After sorting , If arr2 The last value of is less than arr1[x] Then there is no need to compare ,

class Solution {
    //1     2   3   4
    //1     8   9   10
    public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
        Arrays.sort(arr2);
        int count = 0;
        for(int x : arr1){
            int index = binarySearch(arr2,x);
            boolean flag = true;
            if(index<arr2.length){
                flag&=arr2[index]-x>d;
            }
            if(index-1>=0 && index-1<arr2.length){
                flag &= x-arr2[index-1]>d;
            }
            count+=flag?1:0;
        }
        return count;
    }
    private int binarySearch(int[] array,int target){
    
        int left = 0;
        int right = array.length-1;
        while(left<right){
            int middle = left+(right-left)/2;
            if(array[middle]>target){
                right = middle;
            }else{
                left = middle+1;
            }
        }
        return left;
    }
}