Search in Rotated Sorted Array

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.

解题思路：这题旋转排序数组，查找目标值并返回其位置。Rotated：A[i] < A[i + 1] < ... < A[N - 1] > A[0] < A[1] < ... < A[i - 1]。方法采用二分查找法，但是必须注意一些情况。循环遍历，若遇到以下情况(1)当A[middle]>=A[index1]时，判断目标值target是否大于等于A[index1]且小于等于A[middle]，如此index2=middle；否则index1=middle+1(2)A[middle]<A[index1]时，判断target是否小于A[index2]且target是否大于等于middle，则index1=middle;否则index2=middle-1;最后还有一种情况循环结束后，判断A[index2]是否等于target，如此便可以返回目标值的位置。

class Solution {

public:

    int search(int A[], int n, int target) {

        if(A==NULL || n==0)

            return -1;

        int index1=0;

        int index2=n-1;

        int middle;

        while(index1<index2)

        {

            middle=(index1+index2)/2;

            if(A[middle]>=A[index1])

            {

                if(target>=A[index1]&&target<=A[middle])

                    index2=middle;

                else

                    index1=middle+1;

            }

            else

            {

                if(target<=A[index2]&&target>=A[middle])

                    index1=middle;

                else

                    index2=middle-1;

            }

        }

        if(index2>=0 && index2<n && target==A[index2])

            return index2;

        else

            return -1;

    }

};