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.

这道题说如果有重复的元素会不会对，时间之类的有影响，对于我这种懒人直接copy上一题的过去A了，不过感觉和线性搜索O（n）的时间复杂度提升不是太多，只是后面的用了二分查找O（logn）

虽然是一样的，还是贴一下代码，恩，我就是这么无聊

public class Solution {
    public boolean search(int[] A, int target) {
        int i = 0;
        for(; i < A.length; i++){
            if(i > 0 && A[i - 1] > A[i])            //找出轴的位置
                break;
            if(A[i] == target)
                return true;
        }
        if(i == A.length)                            //没有找到
            return false;
        //从轴的位置到最后开始二分查找从i的位置开始到length - 1
        int low = i;
        int high = A.length - 1;
        int middle;
        while(low <= high){
            middle = (low + high) / 2;
            if(A[middle] == target)
                return true;
            else if(A[middle] > target){        //前半段找
                high = middle - 1;
            }else{                                //后半段找
                low = middle + 1;
            }
        }
        
        return  false;
    }
}