81. Search in Rotated Sorted Array II

    /*
     * 81. Search in Rotated Sorted Array II 
     * 2016-5-13 by Mingyang
     * 这就是I的升级版，所有思路一模一样
     * 1.如果右边sort，并且target比中间小，那么肯定在左边，
     * 如果A[mid] == A[high]，那么high--，因为不能判断哪边是sorted
     * 如果A[mid]>A[high]那么左边sort，那么只需要看最左边和target的关系
     * 这里还需要注意的就是while (low <= high)一定需要等于符号
     */
    public boolean search2(int[] A, int target) {
        if (A == null || A.length == 0)
            return false;
        int low = 0;
        int high = A.length - 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (target < A[mid]) {
                if (A[mid] < A[high])// right side is sorted
                    high = mid - 1;// target must in left side
                else if (A[mid] == A[high])
// cannot tell right is sorted, move pointer high----这一部分就是多加出来的
                    high--;
                else // left side is sorted
                  if (target < A[low])
                    low = mid + 1;
                  else
                    high = mid - 1;
            } else if (target > A[mid]) {
                if (A[low] < A[mid])// left side is sorted
                    low = mid + 1;// target must in right side
                else if (A[low] == A[mid])
// cannot tell left is sorted, move pointer low----这一部分就是多加出来的
                    low++;
                else // right side is sorted
                  if (target > A[high])
                    high = mid - 1;
                  else
                    low = mid + 1;
            } else
                return true;
        }
        return false;
    }