LeetCode069 Sqrt(x)

最新推荐文章于 2021-11-30 10:29:11 发布

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本文介绍了一种使用二分查找算法高效计算非负整数平方根的方法，通过不断缩小搜索范围，直到找到最接近目标的整数平方根。文章通过具体示例展示了算法的运行过程。

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.

<思路>尝试过for循环，超时了。所以用二分查找的方式，减少时间。

class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        left,right = 0,x     #二分查找的左右边界

        if 0<x<4:            #0到4区间直接返回1，因为不适用二分查找
            return 1
     
        while True:
            mid = (left+right)//2
            mid2 = mid**2
            if mid2 == x:
                return mid
            if mid2 > x:
                right = mid-1
            elif mid2 < x:
                if (mid+1)**2 > x:
                    return mid
                left = mid+1