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) {…

链接:https://leetcode-cn.com/problems/sqrtx 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. 这道题是一道经典的用二分来解的题,二分有两个模版,当所求的性质在右边的时候,用模版1,当所求性质在左边的…

题目 https://leetcode-cn.com/problems/sqrtx 题解 方法一:牛顿迭代法 按点斜式求出直线方程(即过点Xn,f(Xn)),然后求出直线与x轴交点,即为Xn+1: 求a的平方根即求f(x)=x^2-a的正数解,由牛顿迭代法新一轮解Xn+1=(Xn+a/Xn)/2. 参考链接:https://www.matongxue.com/madocs/205.html 方法二:二分 从0到a/2+1二分查找平方根. 参考链接:https://leetcode-cn.com/…

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…

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