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…

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…

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). 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). 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…

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). 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…

可怕的同时考数值溢出和二分的酱油题之一,常在各种小公司的笔试中充当大题来给你好看... 题意很简单,在<二分查找综述>中有描述. 重点:使用简单粗暴的long long来避免溢出,二分均方根的答案来得到准确解. 当然这里的溢出不止是相乘的溢出,还有第六行那段代码的溢出,每次都会有几个解决问题的斗士牺牲在这里... class Solution { public: int mySqrt(int x) { , b = x; while(a <= b ){ ; if(mid * mid ==…

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…

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…