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

subject ：

 The finger of the sword  Offer 53 - I.  Look up numbers in the sort array  I
 Count the number of times a number appears in the sort array .

 Example  1:

 Input : nums = [5,7,7,8,8,10], target = 8
 Output : 2
 Example  2:

 Input : nums = [5,7,7,8,8,10], target = 6
 Output : 0

 Tips ：

0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums  It is a group of non decreasing numbers 
-109 <= target <= 109

Ideas ： First, determine the left and right boundaries by dichotomy , Then look from left to right .

class Solution {
    
    public int search(int[] nums, int target) {
    
        int n = nums.length;
        int l = 0,r = n - 1;
        int count = 0;

        while (l <= r) {
    
            int mid = (l + r) / 2;
            //  Determine the right boundary 
            if(nums[mid] >= target) {
        
                r = mid - 1;
            }
            //  Determine left boundary 
             else {
    
                 l = mid + 1;
             }
        }
        for( ;l < n && target == nums[l++];){
    
            count++;
        }
        return count;
    }
}

• Time complexity ：O(n)
• Spatial complexity ：O(1)