思路：二分法
时间复杂度：O(logn)
空间复杂度：O(1)

代码：

public class Solution{public int search(int[] nums, int target){int lf=0, rt=n.length-1;while(lf<=rt){int mid = lf+(rt-lf)/2;if(n[mid]>t){rt = mid-1;}else{lf = mid+1;}}return lf;}
}