[LeetCode] Search in Rotated 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.

 

思路：本题思路和http://www.cnblogs.com/vincently/p/4122528.html类似。只是如果有重复的话就不能依靠与边界的比较确定递增的序列。例如，[1,3,1,1,1]，对于A[m]>=A[l]来说，[l,m]的假设就不成立。但是还是可以观察到，左边子数组的值都要大于等于右边子数组的值。将A[m]>=A[l]拆分为两种情况：A[m]>A[l],则区间[l,m]一定递增；如果 A[m]==A[l]确定不了，那就l++，往下看一步。

时间复杂度改变为O(n),空间复杂度O(1)

class Solution {
public:
    bool search(int A[], int n, int target) {
        int first = 0, last = n - 1;
        while (first <= last) {
            const int mid = low + ((high - low) >> 1);
            if (A[mid] == target) return true;
            if (A[first] < A[mid]) {
                if (A[first] <= target && target < A[mid])
                    last = mid - 1;
                else
                    first = mid + 1;
            } else if (A[first] > A[mid]){
                if (A[mid] < target && target <= A[last])
                    first = mid + 1;
                else 
                    last = mid - 1;
            } else {
                first++;
            }
        }
        return false;
    }
};