Implement int sqrt(int x)
.
Compute and return the square root of x.
此题看着简单,实则坑爹啊!!!!!
首先,关于平方,int是不够用的。
其次,二分法肯定好用,但是,最终的结果怎么取?原则上取小于等于的最近整数。但是mid的计算可能会更改这个数,所以需要调整一下。
class Solution { public: int sqrt(int x) { long long left = 0; long long right = x; long long mid = left + (right - left)/2; while(left < right) { if(mid * mid > x) { right = mid -1; } else if(mid * mid < x) { left = mid +1 ; } else return mid; mid = left + (right - left)/2; } if(left * left > x)return left -1; return left; } };