Implement int sqrt(int x)
.
Compute and return the square root of x.
二分可解
这里是关于二分的一些总结:
class Solution { public: int mySqrt(int x) { long long l = 0, r = x,mid; if (x < 2) return x; while (l < r) { mid = l + (r - l + 1) / 2; if (mid > x/mid) r = mid - 1; else if (mid < x/mid) l = mid; else return mid; } return r; } };
原文:http://www.cnblogs.com/pk28/p/7211961.html