#62 Search in Rotated Sorted Array

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题目描述：

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).

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.

Have you met this question in a real interview?

Yes

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

题目思路：

题目要求用O(logN)的时间，明显就是二分法的思路。但是这种rotate sorted的array怎么用binary search呢？我的想法可能有些复杂（因为要用two-pass才解出来），是先用binary search找到整个数组A的最小值的index，假设是sidx，然后我把A再接上一个A，这样从sidx~sidx+A.size()-1这一段坐标，数值都是正常sorted的。这样用传统的二分法就可以找到target了。

这题需要注意的是2点：

1. 用binary search找最小值时，最小值靠左的情况有两种：A[mid] < A[l]（这说明最大值在左边，最小值在包括mid的左边），A[mid] < A[r]（这说明mid往右是正常sorted的）。

2. sidx~sidx+A.size()-1有部分index是超出A.size() - 1的，我用%A.size()就可以得到它在0~A.size()-1的index范围内的相应index。

Mycode（AC = 27ms）：

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) {        // write your code here        if (A.size() == 0) return -1;                // find the index of minimum number in A        int l = 0, r = A.size() - 1, sidx = -1;        while (l + 1 < r) {            int mid = (l + r) / 2; // 2                        if (A[mid] < A[l] || A[mid] < A[r]) {                r = mid;            }            else {                l = mid;            }        }                sidx = A[l] <= A[r]? l : r;                // concatenate 2 A's, so that index start ~ end        // is fully ordered        int start = sidx, end = sidx + A.size() - 1;        while (start <= end) {            int mid = (start + end) / 2;            if (A[mid % A.size()] == target) {                return mid % A.size();            }            else if (A[mid % A.size()] < target) {                start = mid + 1;            }            else {                end = mid - 1;            }        }                return -1;    }};

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