069 Sqrt(x)

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

时间复杂度O(log n )

空间复杂度O(1)

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