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…
69. x 的平方根 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. 示例 1: 输入: 4 输出: 2 示例 2: 输入: 8 输出: 2 说明: 8 的平方根是 2.82842-, 由于返回类型是整数,小数部分将被舍去. class Solution { public int mySqrt(int x) { long t = x; t = 0x5f3759df - (t >> 1)…
69. x 的平方根 非常简单的一个题,用二分法逼近求出ans即可,额外注意下溢出问题. 不过我要给自己增加难度,用long或者BigNum实现没意思,只能使用int类型 换句话当出现溢出时我们自己得检测出来 原代码(会溢出) class Solution { public int mySqrt(int x) { int l = 1; int r = x; while (l < r) { int mid = (l + r) / 2;// l+r会溢出 if (mid * mid == x) {…
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…
更多精彩文章请关注公众号:TanLiuYi00 题目 解题思路 题目要求非负整数 x 的平方根,相当于求函数 y = √x 中 y 的值. 函数 y = √x 图像如下: 从上图中,可以看出函数是单调递增的,满足二分查找的条件(区间是有序的),所以可以用二分查找去做. 解题步骤 比较 mid * mid 跟 x 的大小,相等则直接返回 mid,否则就去以 mid 为分割点的左右区间查找,直到不满足循环循环条件(left == right + 1 或 left == right 究竟是哪一个主要…
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…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:库函数 方法二:牛顿法 方法三:二分查找 日期 题目地址:https://leetcode.com/problems/sqrtx/description/ 题目描述 Implement int sqrt(int x). Compute and return the square root of x. x is guaranteed to be…
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…
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: 4Output: 2Example 2:Input: 8Output: 2Explanation: The square root of 8 is 2.82842..., and since we want to return an integ…
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. 解题思路一: public int mySqrt(int x) { return (int)Math.sqrt(x); } 神奇般的Accepted. 解题思路二: 参考平方根计算方法 计算平方根的算法 这里给出最简单的牛顿法,JAVA实现如下: public int mySqrt(int x) { double g = x; while (Math.abs(g…
Implement int sqrt(int x). Compute and return the square root of x. 原文地址: http://kb.cnblogs.com/page/189867/ 好吧,我承认我标题党了,不过既然你来了,就认真看下去吧,保证你有收获. 我们平时经常会有一些数据运算的操作,需要调用sqrt,exp,abs等函数,那么时候你有没有想过:这个些函数系统是如何实现的?就拿最常用的sqrt函数来说吧,系统怎么来实现这个经常调用的函数呢? 虽然有可能你平…
题目描述: Implement int sqrt(int x). Compute and return the square root of x. 实现开根号,并且返回整数值(这个很重要,不是整数的话就有一种方法用不了了) 方法一:二分法,另外由于我们知道开根号的结果肯定小于等于这个数的二分之一,所以还可以做个线性优化,代码如下: class Solution { public: int sqrt(int x) { ; +; ; while(left<=right){ m=(left+right…
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…
［抄题］: 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. E…
