[sword finger offer] 53 - I. find the number I in the sorted array

The finger of the sword Offer 53 - I. Look up numbers in the sort array I

subject

Count the number of times a number appears in the sort array .

Ideas

  Use binary search to find the left boundary respectively left And the right border right, So the numbers target The number of right - left - 1

Code

class Solution {
    
    public int search(int[] nums, int target) {
    
        //  Search the right border  right
        int i = 0,j = nums.length - 1;
        while(i <= j)
        {
    
            int m = (i + j) / 2;
            if(nums[m] < target) 
            {
    
                i = m + 1;//  explain target In the closed zone [m + 1,j]  On 
            }
            else if(nums[m] ==  target)
            {
    
                i = m + 1;  //  Because I want to find the right boundary   therefore   The right boundary must be in a closed interval [m + 1, j] in 
            }
            else
            {
    
                j = m  -1;// target In the closed zone [i,m - 1] in  
            }
        }
        int right = i; //  Locate right boundary   here right The number pointing to   No target
        if(j >= 0 && nums[j] != target) return 0;

        //  Search left border  right
        i = 0;
        j = nums.length - 1;

        while(i <= j)
        {
    
            int m = (i + j) / 2;
            if(nums[m]  < target)
            {
    
                i = m + 1;
            }
            else{
    
                j = m - 1;
            }
        }
        int  left = j;//  The number that the left boundary points to is not target
        return right - left - 1;
    }
}