[LeetCode-69]-Sqrt(根号运算)

文章目录

题目相关
Solution
- 1. 调用已有的库函数

题目相关

【题目解读】

【原题描述】原题链接
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: 2Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.

【难度】Easy

Solution

1. 调用已有的库函数

#include <math.h>class Solution {public:int mySqrt(int x) {if(x < 0){return 0;}return sqrt(x);}
};

运行效果：

[LeetCode-69]-Sqrt(根号运算)相关推荐

【分治】LeetCode 69. Sqrt(x)
LeetCode 69. Sqrt(x) 参考网址:http://www.cnblogs.com/grandyang/p/4346413.html Solution1: class Solution ...
LeetCode:69. Sqrt(x)
题目链接: 69. Sqrt(x) 题目描述: Implement int sqrt(int x) Compute and return the square root of x. 题目解释: 提干非 ...
LeetCode 69: Sqrt(x) 求根号x(牛顿迭代法和二分查找法)
题目: Implement int sqrt(int x). Compute and return the square root of x. 分析:我们平常可能好少会自己去求解某个数的平方根,一般都 ...
LeetCode 69. Sqrt(x)
题目: Implement int sqrt(int x). Compute and return the square root of x. 思路: 写一个自己的求平方根的函数. 初始化low为0, ...
Leetcode 69 Sqrt(x)
Implement int sqrt(int x). Compute and return the square root of x. 求x的平方根. 二分没什么好说的,注意INT_MAX溢出的情况! ...
LeetCode(69)Sqrt
题目如下: Implement int sqrt(int x). Compute and return the square root of x. 分析如下: (1)借助一个小结论,任何一个数的squ ...
LeetCode | 69. Sqrt(x)
Implement int sqrt(int x). Compute and return the square root of x. 很简单的题: class Solution { public:i ...
LeetCode——[69] Sqrt(x)
int mySqrt(int x) {int i = 1;while(i*i <=x)i++;return --i;} 可把你给厉害坏了.脚趾头想也想到了结局. ✘ Runtime Error✘ ...
LeetCode:69.Sqrt
题目描述: 给你一个非负整数 x ,计算并返回 x 的算术平方根 . 由于返回类型是整数,结果只保留整数部分 ,小数部分将被舍去 . 注意:不允许使用任何内置指数函数和算符,例如 pow(x, ...