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.
题目大意
实现 int sqrt(int x) 函数。计算并返回 x 的平方根,其中 x 是非负整数。由于返回类型是整数,结果只保留整数的部分,小数部分将被舍去。
解题思路
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题目要求求出根号 x
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根据题意,根号 x 的取值范围一定在 [0,x]
之间,这个区间内的值是递增有序的,有边界的,可以用下标访问的,满足这三点正好也就满足了二分搜索的 3 大条件。所以解题思路一,二分搜索。
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解题思路二,牛顿迭代法。求根号 x,即求满足 x^2 - n = 0
方程的所有解。
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