LeetCode Online Judge 题目C# 练习 - Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.

        public static bool SearchinRotatedSortedAarrayII(int[] A, int target)
        {
            return BinarySearchRotatedSortedArrayII(A, 0, A.Length - 1, target);
        }

        public static bool BinarySearchRotatedSortedArrayII(int[] A, int l, int r, int target)
        {
            if (l > r)
                return false;

            int m = (l + r) / 2;

            if (target == A[m])
                return true;

            // if duplicate is allowed, we have to search both upper and lower halves
            // corner case : 8,9,5,8,8,8,8,8 and 8,8,8,8,9,5,8
            if (A[m] == A[l] && A[m] == A[r])
            {
                return BinarySearchRotatedSortedArrayII(A, l, m - 1, target) ||
                        BinarySearchRotatedSortedArrayII(A, m + 1, r, target);
            }

            if (A[m] >= A[l])
            {
                if (target >= A[l] && target < A[m])
                    r = m - 1;
                else
                    l = m + 1;

                return BinarySearchRotatedSortedArrayII(A, l, r, target);
            }
            else
            {
                if (target > A[m] && target <= A[r])
                    l = m + 1;
                else
                    r = m - 1;

                return BinarySearchRotatedSortedArrayII(A, l, r, target);
            }

        }

代码分析：

　　如注释里说的，如果允许重复，那就要前半部后半部都要找。 我承认，我掉入陷阱里了！！！ 如果数组全是 0 ， 要你找1. 那么时间复杂度是多少？O(n)吧。既然是O(n)，干嘛不就loop一遍呢？

        public static bool SearchinRotatedSortedAarrayIIOpt(int[] A, int target)
        {
            for (int i = 0; i < A.Length; i++)
            {
                if (A[i] == target)
                    return true;
            }
            return false;
        }