Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replaceme…

题目: 假设按照升序排序的数组在预先未知的某个点上进行了旋转. ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] ). 搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回 -1 . 你可以假设数组中不存在重复的元素. 你的算法时间复杂度必须是 O(log n) 级别. 示例 1: 输入: nums = [4,5,6,7,0,1,2], target = 0 输出: 4 示例 2: 输入: nums = [4,5,6,7,0,1,2],…

LeetCode 33 Search in Rotated Sorted Array [binary search] <c++> 给出排序好的一维无重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数的下标,如果没有,返回-1.时间复杂度限制\(O(log_2n)\) C++ 我的想法是先找到数组中最大值的位置.然后以此位置将数组一分为二,然后在左右两部分分别寻找target. 二分寻找最大值的时候,因为左半部分的数一定大于nums[l],所以n…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 33: Search in Rotated Sorted Arrayhttps://oj.leetcode.com/problems/search-in-rotated-sorted-array/ Suppose a sorted array is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5…

leetcode - 33. Search in Rotated Sorted Array - Medium descrition Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search.…

33. Search in Rotated Sorted Array Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return i…

这4个题都是针对旋转的排序数组.其中153.154是在旋转的排序数组中找最小值,33.81是在旋转的排序数组中找一个固定的值.且153和33都是没有重复数值的数组,154.81都是针对各自问题的版本1增加了有重复数值的限制条件. 153.154都需要考虑是否旋转成和原数组一样的情况,特别的,154题还需要考虑10111和11101这种特殊情况无法使用二分查找. 33.81则不用考虑这些情况. 找最小值的题主要是利用最小值小于他前一个位置的数,同时也小于后一个位置的数 153. Find Mini…

33. Search in Rotated Sorted Array Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [,,,,,,] might become [,,,,,,]). You are given a target value to search. If found . You may assume no duplicate e…

一.题目说明 这个题目是33. Search in Rotated Sorted Array,说的是在一个"扭转"的有序列表中,查找一个元素,时间复杂度O(logn). 二.我的解答 这是一个查找,根据复杂度,我们知道只能用二分查找.但由于这个不是一个完全的有序列表,故需要改造.先写出二分查找(来源于数据结构): 二分查找: int search(vector<int>& nums, int target){ if(nums.size()<0) return…

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, otherwise return -1. You…