leetcode算法题----Sqrt(x)

题目:69. Sqrt(x)

Easy

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.

求x的平方根，返回小于等于实际平方根的整数。

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.

解题思路：求x的平方根，从0到x，依次判断平方与x的大小关系。即从x+1个数中查找一个数。采用二分查找的方法。

c代码实现：

int mySqrt(int x) {
    if (x<=1)
        return x;
    int left = 0;
    int right = x;
    int mid;
    while(left=mid){
            left = mid+1;
        }
        else{
            right=mid;
        }
    }
    return right-1; 
}

python代码实现：

class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        if(x<2):
            return x
        left = 0
        right = x
        while(left=mid):
                left=mid+1
            else:
                right=mid
        return right-1

参考：

[LeetCode] Sqrt(x) 求平方根    https://www.cnblogs.com/grandyang/p/4346413.html