LeetCode 69 Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

牛顿迭代法：关于牛顿迭代法，详解http://blog.youkuaiyun.com/niuooniuoo/article/details/51787807

	public int mySqrt(int x) {
		if (x == 0) return 0;
		double last = 0.0;
		double res = 1.0;
		while (res != last) {
			last = res;
			res = (res + x / res) / 2;
		}
		return (int) res;
	}

二分查找法

	int sqrt(int x) {
		long i = 0, j = x / 2 + 1;
		while (i <= j) {
			long mid = (i + j) / 2;
			long sq = mid * mid;
			if (sq == x) return (int) mid;
			else if (sq < x) i = mid + 1;
			else j = mid - 1;
		}
		return (int) j;
	}

详解参考：http://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html