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…

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] -> …

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…

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 ||…

题目: 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 du…

题目: 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. 题解: class Solution { public: /** * @param nums…

LintCode 160. Find Minimum in Rotated Sorted Array II (Medium) LeetCode 154. Find Minimum in Rotated Sorted Array II (Hard) 解法1 自己做了半个小时, 想分情况判断num[mid]与num[start], num[mid]与num[end]的大小关系, 每个关系分>, =, <的情况, 于是就是9种情况... >, >, L = M + 1 >, =,…

LintCode 159. Find Minimum in Rotated Sorted Array (Medium) LeetCode 153. Find Minimum in Rotated Sorted Array (Medium) 这题看着简单, 但是条件总容易搞错. @_@... 思路是在当前区间有序的时候立即停止, 然后某个点(详见代码)就是答案. 我没有做nums.size() == 0的判断, 因为这种情况应该抛个异常什么的, 给什么int值都可能是有效值(如果nums中的数没有…

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…