Given a rotated sorted array, recover it to sorted array in-place. Clarification What is rotated array? For example, the orginal array is [1,2,3,4], The rotated array of it can be [1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3] Example [4, 5, 1, 2, 3] -> …

Given a rotated sorted array, recover it to sorted array in-place. Example [4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5] Challenge In-place, O(1) extra space and O(n) time. Clarification What is rotated array: - For example, the orginal array is [1,2,3,4], The…

题目: 恢复旋转排序数组 给定一个旋转排序数组,在原地恢复其排序. 样例 [4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5] 挑战 使用O(1)的额外空间和O(n)时间复杂度 说明 什么是旋转数组? 比如,原始数组为[1,2,3,4], 则其旋转数组可以是[1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3] 解题: 开始我想,先找到中间的临界点,然后在排序,临界点找到了,排序不知道怎么搞了,在这里,看到了很好的方法,前半部分逆序,后半部分逆序,整…

一句话思路:从左边开始的三步翻转法 一刷报错: 不理解start.end是位置随机定义的.i,j是临时变量,为start,end服务 nums.size()区别于nums.length:用于范形变量.作用于一堆.但是如果都是从0开始,-1的原理相同 index可以在括号里定义 if (nums.get(index) > nums.get(index + 1)),不是index++,前面也会变 set是一个方法,不能直接用set,必须用nums.set index指向的前一个元素比较大的时候,才要…

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be…

Suppose a sorted array 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). Find the minimum element. You may assume no duplicate exists in the array. 这道寻找旋转有序数组的最小值肯定不能通过直接遍历整个数组来寻找,这个方法过于简单粗暴,这样的话,旋不…

Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 这道是之前那道Search in Rotated Sorted Array 在旋转有序数组…

Suppose a sorted array 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 may assume no duplic…

题目概述: Suppose a sorted array 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).Find the minimum element.You may assume no duplicate exists in the array. 解题思路: 在一个有顺序的数组中查找个最小值,很容易想到的就是二分法,的确,这里用的就是二分,…

1. 画图, 直观. 2. 讨论数组为空或者个数为零. 3. 讨论首尾, 若为翻转过的则进行查找直到最后两个数进行比较, 取小者. public class Solution { /** * @param num: a rotated sorted array * @return: the minimum number in the array */ public int findMin(int[] num) { // write your code here if(num == null ||…