LeetCode-154 Find Minimum in Rotated Sorted Array II

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

 

思路：

设数组num, 左边起点为l,右边终点为r, m=(l+r)/2

1. 如果num[l] < num[r];说明从l到r有序，则min=num[l];

2. 如果num[l] > num[r];说明发生了折叠；

　　若num[m]<=num[r];则在l~m间查找最小值；

　　若num[m]>=num[l];则在m+1~r间查找最小值；

3. 如果num[l] == num[r];

　　若num[m] < num[r];在l~m间查找最小值；

　　若num[m] > num[l];在m+1~r间查找最小值

　　若num[m] == num[l];此时二分查找失效，需要遍历从l~r整个数组（如[3,3,3,3,3,3,3,1,3]和[3,3,1,3,3,3,3,3,3]）

代码如下：

public int findMin(int[] num) {
        return findMin(num, 0, num.length-1);
    }
    
    public int findMin(int[] num ,int l, int r) {
        if(l == r)
            return num[l];
        if(l > r)
            return 0;
        int m = (l+r) / 2;
        if(num[l] == num[r] && num[r] == num[m]) {
            int min = num[l];
            for(int i=l+1; i<=r; i++) {
                if(num[i] < min)
                    min = num[i];
            }
            return min;
        } else if(num[l] < num[r]) {
            return num[l];
        } else {
            if(num[m] <= num[r]) 
                return findMin(num, l, m);
            else if(num[m] >= num[l])
                return findMin(num, m+1, r);
        }
        return 0;
    }