LeetCode 69. Sqrt(x)，求根算法

69. Sqrt(x)

Implement 

int sqrt(int x)

.

Compute and return the square root of x.

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这道题要找x的平方根，x的平方根肯定小于x/2。要在[1,x/2]有序序列当中找一个数，用二分法：

public int mySqrt(int x) {
long high = (x / 2) + 1;
long low = 0;
while (high >= low) {
long mid = (high + low) / 2;
if (mid * mid == x)
return (int)mid;
else if (mid * mid > x)
high = mid - 1;
else
low = mid + 1;
}
return (int)high;
}