Title Description

Ideas ： find target-1 The rightmost sum of target The rightmost number , Two less is good

The code is as follows ：

class Solution {
    public int search(int[] nums, int target) {
        // Find the rightmost number , Subtraction of two numbers 
        //target-1  The rightmost number ,target The rightmost number 
        return binarysearch(nums,target)-binarysearch(nums,target-1);

    }
    public int binarysearch(int[] nums,int target)
    {
        int left=0;
        int right=nums.length;
        while(left<right)
        {
            int mid=left+(right-left)/2;//right=n Guaranteed binary sum left<right
            if(nums[mid]>target)
            {
                right=mid;
            }
            else//<= All to the left 
            {
                left=mid+1;
            }
        }
        return left;
    }
}

Possible doubts

1. How to ensure to find the rightmost number ？

while(left < right)

------------------------
     else//<= All to the left 
            {
                left=mid+1;
            }

 These two paragraphs realize finding the rightmost number 

 give an example : When the result is [4,5] ,  The goal is 4 when ,
left = 4
right = 5
mid = 4
 According to judgment ,left = mid +1;
 here ,left == right Out of the loop 
left-1  perhaps  right-1  Namely target  The rightmost number of 

To find the rightmost number

2.right = nums.length instead of   nums.length-1

1. because nums[right]  There was no access to 
2. To ensure the mid  Half every time 
int mid=left+(right-left)/2

3.mid = left +(right-left) /2 instead of mid = (left+right)/2

When left and right When I was younger , Both cases are ok

But one situation is left and right They're all bigger , here ,left + right Will cause removal , So subtraction is better