Search in Rotated Sorted Array

Description

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e.,0 1 2 4 5 6 7might become4 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.

Example

For[4, 5, 1, 2, 3]andtarget=1, return2.

For[4, 5, 1, 2, 3]andtarget=0, return-1.

Challenge

O(logN) time

Related problems

Search in Rotated Sorted Array II

Search a 2D Matrix

Implementation

Link: http://lintcode.com/en/problem/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) {
        // write your code here
        if(A.size() == 0) return -1;

        int start = 0, end = A.size() - 1;

        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(A[mid] == target) {
                return mid;
            } else if(A[mid] < A[end]) {
                if(A[mid] < target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            } else {
                if(A[mid] > target && target >= A[start]) {
                    end = mid;
                } else {
                    start = mid;
                }
            }
        }
        if(A[start] == target) return start;
        if(A[end] <= target) return end;
        return -1;
    }
};