leetcode009--用二分查找在数组中搜索目标值

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

public class test09 {
    /**
     *
     给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
     请必须使用时间复杂度为 O(log n) 的算法。

     示例 1:
     输入: nums = [1,3,5,6], target = 5
     输出: 2

     示例 2:
     输入: nums = [1,3,5,6], target = 2
     输出: 1

     示例 3:
     输入: nums = [1,3,5,6], target = 7
     输出: 4

     示例 4:
     输入: nums = [1,3,5,6], target = 0
     输出: 0

     示例 5:
     输入: nums = [1], target = 0
     输出: 0

     提示:
     1 <= nums.length <= 104
     -104 <= nums[i] <= 104
     nums 为无重复元素的升序排列数组
     -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;
    }
}