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…
A peak element is an element that is greater than its neighbors. Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the pea…
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1].…
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…
You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad ve…
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…