# Search in Rotated Sorted Array ### Source - leetcode: [Search in Rotated Sorted Array | LeetCode OJ](https://leetcode.com/problems/search-in-rotated-sorted-array/) - lintcode: [(62) Search in Rotated Sorted Array](http://www.lintcode.com/en/problem/search-in-rotated-sorted-array/) ### Problem 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 itsindex, otherwise return -1. You may assume no duplicate exists in the array. #### Example For `[4, 5, 1, 2, 3]` and `target=1`, return `2`. For `[4, 5, 1, 2, 3]` and `target=0`, return `-1`. #### Challenge O(logN) time ### 题解 - 找到有序数组 对于旋转数组的分析可使用画图的方法，如下图所示，升序数组经旋转后可能为如下两种形式。  对于有序数组，使用二分搜索比较方便。分析题中的数组特点，旋转后初看是乱序数组，但仔细一看其实里面是存在两段有序数组的。刚开始做这道题时可能会去比较`target`和`A[mid]`, 但分析起来异常复杂。**该题较为巧妙的地方在于如何找出旋转数组中的局部有序数组，并使用二分搜索解之。**结合实际数组在纸上分析较为方便。 ### C++ ~~~ /** * 本代码fork自 * http://www.jiuzhang.com/solutions/search-in-rotated-sorted-array/ */ class Solution { /** * param A : an integer ratated sorted array * param target : an integer to be searched * return : an integer */ public: int search(vector<int> &A, int target) { if (A.empty()) { return -1; } vector<int>::size_type start = 0; vector<int>::size_type end = A.size() - 1; vector<int>::size_type mid; while (start + 1 < end) { mid = start + (end - start) / 2; if (target == A[mid]) { return mid; } if (A[start] < A[mid]) { // situation 1, numbers between start and mid are sorted if (A[start] <= target && target < A[mid]) { end = mid; } else { start = mid; } } else { // situation 2, numbers between mid and end are sorted if (A[mid] < target && target <= A[end]) { start = mid; } else { end = mid; } } } if (A[start] == target) { return start; } if (A[end] == target) { return end; } return -1; } }; ~~~ ### Java ~~~ public class Solution { /** *@param A : an integer rotated sorted array *@param target : an integer to be searched *return : an integer */ public int search(int[] A, int target) { if (A == null || A.length == 0) return -1; int lb = 0, ub = A.length - 1; while (lb + 1 < ub) { int mid = lb + (ub - lb) / 2; if (A[mid] == target) return mid; if (A[mid] > A[lb]) { // case1: numbers between lb and mid are sorted if (A[lb] <= target && target <= A[mid]) { ub = mid; } else { lb = mid; } } else { // case2: numbers between mid and ub are sorted if (A[mid] <= target && target <= A[ub]) { lb = mid; } else { ub = mid; } } } if (A[lb] == target) { return lb; } else if (A[ub] == target) { return ub; } return -1; } } ~~~ ### 源码分析 1. 若`target == A[mid]`，索引找到，直接返回 1. 寻找局部有序数组，分析`A[mid]`和两段有序的数组特点，由于旋转后前面有序数组最小值都比后面有序数组最大值大。故若`A[start] < A[mid]`成立，则start与mid间的元素必有序（要么是前一段有序数组，要么是后一段有序数组，还有可能是未旋转数组）。 1. 接着在有序数组`A[start]~A[mid]`间进行二分搜索，但能在`A[start]~A[mid]`间搜索的前提是`A[start] <= target <= A[mid]`。 1. 接着在有序数组`A[mid]~A[end]`间进行二分搜索，注意前提条件。 1. 搜索完毕时索引若不是mid或者未满足while循环条件，则测试A[start]或者A[end]是否满足条件。 1. 最后若未找到满足条件的索引，则返回-1. ### 复杂度分析 分两段二分，时间复杂度仍近似为 O(logn)O(\log n)O(logn).