LeetCode 69. Sqrt(x)

原题链接在这里：https://leetcode.com/problems/sqrtx/

题目：

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 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

题解：

先找middle = (left + right)/2, middle^2 和 x 比较，比x小就在middle 和 right这段找，反之亦然。

Note:1. middle 要设为long, 否则middle * middle 可能overflow

2. while loop的条件是l<=r, 不是< e.g. x = 2

3. return 别忘了加cast

4. 即使这不是一个perfect square 也是返回right, 因为此时right就是那个较小的值.

Time Complexity: O(logx). Space: O(1).

AC Java:

public class Solution {
    public int mySqrt(int x) {
        if(x<0){
            throw new IllegalArgumentException("Invalid input.");
        }
        
        long l = 0; 
        long r = x;
        while(l<=r){
            long mid = l+(r-l)/2;
            if(mid*mid > x){
                r = mid-1;
            }else if(mid*mid < x){
                l = mid+1;
            }else{
                return (int)mid;
            }
        }
        return (int)r;
    }
}

类似与Pow(x, n), Valid Perfect Square.