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…
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…
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…
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].…
题意:找到第一个出问题的版本 二分查找,注意 mid = l + (r - l + 1) / 2;因为整数会溢出 // Forward declaration of isBadVersion API. bool isBadVersion(int version); class Solution { public: int firstBadVersion(int n) { , r = n , ans ; while(l <= r){ ) / ; if(!isBadVersion(mid)){ l…
题意: 有一个bool序列表示对应下标的版本是否出问题(下标从1开始),如果一个版本出了问题,那么其后面全部版本必定出问题.现在给出判断任意版本是否出问题的API,请找到第一个出问题的版本. 思路: 明显的二分查找. // Forward declaration of isBadVersion API. bool isBadVersion(int version); class Solution { public: int firstBadVersion(int n) { , R=n; whil…
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…
二分查找又叫折半查找,要查找的前提是检索结果位于已排序的列表中. 概念 在一个已排序的数组seq中,使用二分查找v,假如这个数组的范围是[low...high],我们要的v就在这个范围里.查找的方法是拿low到high的正中间的值,我们假设是m,来跟v相比,如果m>v,说明我们要查找的v在前数组seq的前半部,否则就在后半部.无论是在前半部还是后半部,将那部分再次折半查找,重复这个过程,知道查找到v值所在的地方. 实现二分查找可以用循环,也可以递归,先给出两种方式的伪代码. 伪代码 使用循环实现…
二分查找的思想很简单,它是针对于有序数组的,相当于数组(设为int a[N])排成一颗二叉平衡树(左子节点<=父节点<=右子节点),然后从根节点(对应数组下标a[N/2])开始判断,若值<=当前节点则到左子树,否则到右子树.查找时间复杂度是O(logN),因为树的高度是logN. 二分查找分两种:一种是精确查找某个元素x,另一种则是根据比较关系式查找,比如返回i使得对任意j<i均有a[j]<a[i],这用在折半插入排序中. 刚好前天无聊不借助STL手写折半插入排序时,发现自己…
如题 (总结) -认真读题, 还WA了一次, https://www.nowcoder.com/practice/abc3fe2ce8e146608e868a70efebf62e?tpId=13&tqId=11154&tPage=1&rp=1&ru=%2Fta%2Fcoding-interviews&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking 这题也没有说数据量,暂且二分吧. 借鉴学习文章列表 链接1: 链接2:…