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…

69. Sqrt(x) Total Accepted: 93296 Total Submissions: 368340 Difficulty: Medium 提交网址: https://leetcode.com/problems/sqrtx/ Implement int sqrt(int x). Compute and return the square root of x. 分析: 解法1:牛顿迭代法(牛顿切线法) Newton's Method(牛顿切线法)是由艾萨克·牛顿在<流数法>(M…

Leetcode 69. Sqrt(x) Easy https://leetcode.com/problems/sqrtx/ 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 truncat…

Question 69. Sqrt(x) Solution 题目大意: 求一个数的平方根 思路: 二分查找 Python实现: def sqrt(x): l = 0 r = x + 1 while l < r: m = l + (r - l) // 2 if m * m > x: r = m else: l = m + 1 return l - 1…

这是悦乐书的第158次更新,第160篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第17题(顺位题号是69). 计算并返回x的平方根,其中x保证为非负整数. 由于返回类型是整数,因此将截断十进制数字,并仅返回结果的整数部分.例如: 输入:4 输出:2 输入:8 输出:2 说明:8的平方根是2.82842 ...,从2以后小数部分被截断,返回2 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试.…

题目: Implement int sqrt(int x). Compute and return the square root of x. 链接:   http://leetcode.com/problems/sqrtx/ 题解: 求平方根. 二分法, Time Complexity - O(logn), Space Complexity - O(1) public class Solution { public int mySqrt(int x) { if(x <= 1) return x…

Implement int sqrt(int x). Compute and return the square root of x. x is guaranteed to be a non-negative integer. Example 1: Input: 4 Output: 2 Example 2: Input: 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since we want to return…

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…

Implement int sqrt(int x). 思路: Binary Search class Solution(object): def mySqrt(self, x): """ :type x: int :rtype: int """ l = 0 r = x while l <= r: mid = (l+r)//2 if mid*mid < x: l = mid + 1 elif mid*mid > x: r = mi…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Implement int sqrt(int x). Compute and return the square root of x. (二)解题 实现sqrt(x),找到一个数,它的平方等于小于x的最接近x的数. class Solution { public: int mySqrt(int x) { int…