[leetcode] 69. Sqrt(x) 解题报告

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Leetcode 69. Sqrt(x) 解题报告【C库函数sqrt(x)模拟-求平方根】

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

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Pascal's Triangle Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] SOLUTION 1:很easy的题.注意记得把List加到ret中.比较简单,每一行的每一个元素有这个规律:1. 左右2边的是1.i, j 表示行,列坐标.2.

LeetCode --- 69. Sqrt(x)

题目链接:Sqrt(x) Implement int sqrt(int x). Compute and return the square root of x. 这道题的要求是实现int sqrt(int x),即计算x的平方根. 考虑二分,即先令l和r分别为1和x/2+1(x的平方根一定小于等于x/2+1),然后m等于(l+r)/2,不断比较m*m和x的大小. 由于m*m的时候,可能溢出,因此可以用除法代替乘法,或者采用long long类型. 时间复杂度:O(logn) 空间复杂度:O(1)

[LeetCode]Longest Valid Parentheses, 解题报告

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Leetcode 69. Sqrt(x)及其扩展(有/无精度、二分法、牛顿法)详解

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

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