【Leetcode】Search in Rotated Sorted Array II

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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.

思路：跟【Leetcode】Search in Rotated Sorted Array類似，但是需要做些許改動，因為對於遞增序列的判斷需要增加一些條件。當然，此題的解答也能解決上一題的問題。

代碼一：

class Solution {public:    bool search(int A[], int n, int target) {        if(n <= 0)  return false;                int left = 0;        int right = n - 1;        int middle = 0;                while(left <= right)        {            middle = (left + right) / 2;                        if(A[middle] == target)                return true;            else if(A[middle] < target)            {                if(A[left] == A[middle])                        left++;                else if(A[left] > A[middle] && A[right] < target)                    right = middle - 1;                else                    left = middle + 1;            }            else            {                if(A[right] == A[middle])                       right--;                else if(A[right] < A[middle] && A[left] > target)                    left = middle + 1;                else                    right = middle - 1;            }        }                return false;    }};

代碼二：

class Solution {public:    bool search(int A[], int n, int target) {        if(n <= 0)  return false;                int left = 0;        int right = n - 1;        int middle = 0;                while(left <= right)        {            middle = (left + right) / 2;                        if(A[middle] == target)                return true;                        if(A[left] < A[middle])            {                if(A[left] <= target && target < A[middle])                    right = middle - 1;                else                    left = middle + 1;            }            else if(A[left] > A[middle])            {                if(A[middle] < target && target <= A[right])                    left = middle + 1;                else                    right = middle - 1;            }            else                left++;        }                return false;    }};