leetcode之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.

这道题和Search in Rotated Sorted Array I不同之处在于数组中有重复数字,因此用二分查找的时候要注意一些分界点的检查。这道题我没有自己去写,而是从网上找的别人的源码。

public boolean search(int[] A, int target) {
        int begin = 0;  
        int end = A.length-1;  
        while(begin < end)  
        {  
            int mid = (begin+end)/2;  
            if(A[mid] == target)return true;  
            else if(A[begin] == A[mid])  
            {  
                for(int i = begin; i< mid; i++)  
                    if(A[i]==target)return true;  
                begin = mid+1;  
            }  
            else if(A[begin] < A[mid])//begin-mid increase  
            {  
                if(A[begin] <= target && target < A[mid])  
                    end = mid-1;  
                else  
                    begin = mid+1;  
            }  
            else//mid-end increase  
            {  
                if(A[mid] < target && target <= A[end])  
                    begin = mid+1;  
                else  
                    end = mid-1;  
            }  
        }  
        if(begin==end && A[begin]==target)  
            return true;  
        else  
            return false; 
        
    }

  

原文地址:https://www.cnblogs.com/gracyandjohn/p/4409072.html