141. Sqrt(x) 【easy】

141. Sqrt(x) 【easy】

Implement int sqrt(int x).

Compute and return the square root of x.

Example

sqrt(3) = 1

sqrt(4) = 2

sqrt(5) = 2

sqrt(10) = 3

Challenge 

O(log(x))

解法一：

class Solution {
public:
    /*
     * @param x: An integer
     * @return: The sqrt of x
     */
    int sqrt(int x) {
        // write your code here
        long start = 1;
        long end = x;
        
        while (start + 1 < end) {
            long mid = start + (end - start) / 2;
            
            if (mid * mid == x) {
                return mid;
            }
            else if (mid * mid < x) {
                start = mid;
            }
            else if (mid * mid > x) {
                end = mid;
            }
        }
        
        if (end * end <= x) {
            return end;
        }
        else {
            return start;
        }
    }
};

注意：

1、start、end、mid用long类型防止溢出

2、最后判断end * end与x的关系时要考虑等号，否则输入数值为0的时候就错了。

解法二：

class Solution {
public:
    /**
     * @param x: An integer
     * @return: The sqrt of x
     */
    int sqrt(int x) {
        // write your code here
        long left = 0;
        if (x == 1)
            return 1;
        long right = x;
        long mid = left + (right - left) / 2;
        while (left + 1 < right) {
            if (x > mid * mid) {
                left = mid;
            } else if (x < mid * mid) {
                right = mid;
            } else {
                return mid;
            }
            mid = left + (right - left) / 2;  
        }
        return left;
    }
};

这里还是两头都判断一下比较保险。