Implement
int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example：
Input: 4
Output: 2
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
解释下题目：
实现开平方根的函数
1. 二分法
实际耗时：22ms
public int mySqrt(int x) {
if (0 == x) {
return 0;
}
int small = 0;
int big = Integer.MAX_VALUE;
while (true) {
int mid = small + ((big - small) >> 1);
if (mid > x / mid) {
big = mid - 1;
} else if (mid + 1 > x / (mid + 1)) {
return mid;
} else {
small = mid + 1;
}
}
}
思路没什么好说的，就是二分法判断呗
