[LeetCode] 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.

分析
如果允许重复元素，则Search in Rotated Sorted Array题中如果 A[m]>=A[l], 那么 [l,m] 为递增序列的假设就不能成立了，比
如 [1,2，3,1,1,1]。
如果 A[m]>=A[l] 不能确定递增，那就把它拆分成两个条件：
• 若 A[m]>A[l]，则区间 [l,m] 一定递增
• 若 A[m]==A[l] 确定不了，那就 l++，往下看一步即可。

Search in Rotated Sorted Array 的code：

class Solution {
public:
    bool search(int A[], int n, int target) {
        int low = 0;
        int high = n-1;
        int mid ;

        
        while(low<=high)
        {
            mid= (low + high)/2;
            
            if(A[mid]==target)
            {
                return true;
            }
                    
            if(A[low]<=A[mid]) // the pivot is in the bottom half
            {
                if(A[low]<=target && target < A[mid]) //target in the first half
                    high = mid-1;
                else
                    low = mid+1;//target in the bottom half
            }
            else // the pivot is in the first half
            {
                if(A[mid] < target && target <= A[high])// the pivot is in the first half
                    low = mid+1;
                else //target in the first half
                    high = mid-1;
            }
            
        }
        return false;
        
    }
};

将if (A[low] <= A[mid]) 拆成if (A[low] < A[mid]) 和 if (A[low] == A[mid]) 两个分支，code如下：

class Solution {
public:
    bool search(int A[], int n, int target) {
        int low = 0;
        int high = n-1;
        int mid ;

        
        while(low<=high)
        {
            mid= (low + high)/2;
            
            if(A[mid]==target)
            {
                return true;
            }
                    
            if(A[low]<A[mid])
            {
                if(A[low]<=target && target < A[mid])
                    high = mid-1;
                else
                    low = mid+1;
            }
            else if(A[low] > A[mid])
            {
                if(A[mid] < target && target <= A[high])
                    low = mid+1;
                else 
                    high = mid-1;
            }
            else 
                low++;
            
        }
        return false;
        
    }
};