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;
}
}