Search in Rotated Sorted Array

Q:

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.

还是使用二分法，通过比较最左边，中间以及target的大小关系可以确认是向左走或向右走。当然，这是数组中无重复的情形，如果有重复还需要考虑的更加完善。

A:

class Solution {
public:
    int search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (n < 0) return -1;
        int begin = 0;
        int end = n - 1;
        
        while (begin <= end) {
            int pos = (begin + end) / 2;
            if (A[pos] == target) {
                return pos;
            }
            if (A[begin] == target) return begin;
            
            if (A[pos] < target) {
                if (A[begin] > target) {
                    begin = pos + 1;
                } else {
                    if (A[begin] <= A[pos]) {
                        begin = pos + 1;
                    } else {
                        end = pos - 1;
                    }
                }
            } else {
                if (A[begin] < target) {
                    end = pos - 1;
                } else {
                    if (A[begin] <= A[pos]) {
                        begin = pos + 1;
                    } else {
                        end = pos - 1;
                    }
                }
            }
        }
        return -1;
    }
};