leetcode解题报告（4）：Search in Rotated Sorted ArrayII

描述

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，那么上一题中“若A[first] <= A[mid]，则[first,mid]区间有序”的假设就不成立了，比如可能存在旋转后形如[1,3,1,1,1]的数组。

显然，如果A[first]<=A[mid]不能确定是否有序，就把这个条件细分一下：

• 若A[first] < A[mid]，就一定递增；
• 若A[first] == A[mid]，则确定不了，就将first加1，往下看一步即可。

代码如下：

class Solution{
public:
    int search(int A[],int n,int target){
        int first = 0,last = n;
        const int mid = (first + mid) / 2;  //note that mid is const,which means that we would not modify the value of mid before a new loop starts
        while(first != mid){
            if(A[mid] == target)
                return mid;
            if(A[first] < A[mid]){
                if(A[first] <= target && target < A[mid])
                    last = mid;
                else
                    first = mid + 1;
            }
            else if(A[mid] == A[first])
                ++first;    //skip duplicatted one
            else{
                if(A[mid] <= target && target <= A[last - 1])
                    first = mid + 1;
                else
                    last = mid;
            }
        }
        return -1;
    }
}