说明

算法：Search in Rotated Sorted Array
地址：https://leetcode.com/problems/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 its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

解题思路

注意：要用二分查找。既然有部分数组被翻转了，找到被翻转的数据(叫做divideIndex)，就可以确定target在哪个部分，用二分查找即可。
比如 Example 1： [4,5,6,7,0,1,2]，

找到分割的数据0，index = 4;
判断 nums[index] 刚好等于 0, 直接return index.
如果不等于，比如Example 3, target = 3, 跟nums[lastIndex]比较。
发现3 > 2, 所以，数据在范围[0, divideIndex - 1]区间查找.
最后如果小于，发现3 < ?, 数据在范围[divideIndex + 1, lastIndex] 中查找。

注意：要考虑到边界的问题，比如数组是空的情况，直接返回-1.

代码如下：

public int search(int[] nums, int target) {// check edgeif (nums == null || nums.length == 0) {return -1;}// find the divide indexint divideIndex = findDividIndex(nums);if (target == nums[divideIndex]) {return divideIndex;}int lastIndex = nums.length - 1;int beginIndex = target > nums[lastIndex] ? 0 : divideIndex;int endIndex = target > nums[lastIndex] ? divideIndex : lastIndex;while (beginIndex <= endIndex) {int midIndex = beginIndex + ((endIndex - beginIndex) >> 1);if (target == nums[midIndex]) {return midIndex;} else if (target > nums[midIndex]) {beginIndex = midIndex + 1;} else {endIndex = midIndex - 1;}}return -1;}private int findDividIndex(int[] nums) {int from = 0;int to = nums.length - 1;while (from < to) {int mid = from + ((to - from) >> 1);if (nums[mid] > nums[to]) {from = mid + 1;} else {to = mid;}}return from;}

代码下载：

https://github.com/zgpeace/awesome-java-leetcode/blob/master/code/LeetCode/src/binarysearch/SearchInRotatedSortedArray.java

算法:Search in Rotated Sorted Array(搜索部分顺序被翻转的数组)相关推荐

python旋转排序数组_LeetCode（力扣）——Search in Rotated Sorted Array 搜索旋转排序数组 python实现...
题目描述: python实现Search in Rotated Sorted Array 搜索旋转排序数组中文: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, ...
LeetCode Search in Rotated Sorted Array II -- 有重复的旋转序列搜索
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this ...
[算法题] Search in Rotated Sorted Array ii
题目内容题目来源:LeetCode Suppose an array sorted in ascending order is rotated at some pivot unknown to yo ...
81. Search in Rotated Sorted Array II
题目: Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would ...
【leetcode】Search in Rotated Sorted Array II（middle）☆
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this ...
LeetCode Search in Rotated Sorted Array II
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would t ...
62. Search in Rotated Sorted Array【medium】
62. Search in Rotated Sorted Array[medium] Suppose a sorted array is rotated at some pivot unknown t ...
leetcode 81 Search in Rotated Sorted Array II ----- java
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this ...
Search in Rotated Sorted Array II
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this ...
LeetCode Online Judge 题目C# 练习 - Search in Rotated Sorted Array II
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this ...

算法:Search in Rotated Sorted Array(搜索部分顺序被翻转的数组)

说明

解题思路

代码如下：

代码下载：

算法:Search in Rotated Sorted Array(搜索部分顺序被翻转的数组)相关推荐

最新文章

热门文章