【leetcode】Search in Rotated Sorted Array II

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.

 

与I类似，只是在二分搜索前，只是在判断左右两边是否有序时更加麻烦一些。

此时复杂度是O(n)了，和采用线性搜索的方法基本一致

 

class Solution {
public:
    bool search(int A[], int n, int target) {
       
       
        int left=0;
        int right=n-1;
        int mid;
        while(left<=right)
        {
           
            mid=(left+right)/2;
            if(A[mid]==target)
            {
                return true;
            }
           
 
            //增加了这些判断语句
            bool isLeft=true;
           
            for(int i=left;i<=mid;i++)
            {
                if(A[i]>A[mid])
                {
                    isLeft=false;
                    break;
                }
            }
           
            if(isLeft)
            {
                if(A[left]<=target&&target<A[mid])
                {
                    right=mid-1;
                }
                else
                {
                    left=mid+1;
                }
            }
            else
            {
                if(A[mid]<target&&target<=A[right])
                {
                    left=mid+1;
                }
                else
                {
                    right=mid-1;
                }
            }
 
        }
        return false;
       
    }
};