Leetcode009 -- search the target value in the array with binary search

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

public class test09 {
    /**
     *
      Given a sort array and a target value , Find the target value in the array , And return its index . If the target value does not exist in the array , Return to where it will be inserted in sequence .
      Please use a time complexity of  O(log n)  The algorithm of .

      Example  1:
      Input : nums = [1,3,5,6], target = 5
      Output : 2

      Example  2:
      Input : nums = [1,3,5,6], target = 2
      Output : 1

      Example  3:
      Input : nums = [1,3,5,6], target = 7
      Output : 4

      Example  4:
      Input : nums = [1,3,5,6], target = 0
      Output : 0

      Example  5:
      Input : nums = [1], target = 0
      Output : 0

      Tips :
     1 <= nums.length <= 104
     -104 <= nums[i] <= 104
     nums  Arrange the array in ascending order without repeating elements 
     -104 <= target <= 104
     */
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] nums0 = br.readLine().split(",");
        String target0 = br.readLine();
        int target = Integer.parseInt(target0);
        int[] nums = new int[nums0.length];
        int i = 0;
        for(String num: nums0){
            nums[i++] = Integer.parseInt(num);
        }

        System.out.println(searchInsert(nums,target));

    }

    public static int searchInsert(int[] nums, int target){
        int n = nums.length;
        int left = 0, right = n - 1, ans = n;
        while (left <= right) {
            int mid = ((right - left) / 2) + left;
            if (target <= nums[mid]) {
                ans = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
}