[LeetCode] 154. Find Minimum in Rotated Sorted Array II 寻找旋转有序数组的最小值 II

Posted 2020-10-26 轻风舞动

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.

153. Find Minimum in Rotated Sorted Array 的拓展，这个题里允许有重复的元素。原来是依靠中间和边缘元素的大小关系，来判断哪一半是有序的。而现在因为重复元素的出现，如果遇到中间和边缘相等的情况，就无法判断哪边有序，因为哪边都有可能有序。假设原数组是{1,2,3,3,3,3,3}，那么旋转之后有可能是{3,3,3,3,3,1,2}，或者{3,1,2,3,3,3,3}，判断左边缘和中心的时候都是3，就不知道应该截掉哪一半。解决的办法是对边缘移动一步，直到边缘和中间不在相等或者相遇，这就导致了会有不能切去一半的可能。所以最坏情况就会出现每次移动一步，总共移动n，算法的时间复杂度j: O(logn) ~ O(n)。

Java:

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

Python:

class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        left, right = 0, len(nums) - 1
        while left < right:
            mid = left + (right - left) / 2

            if nums[mid] == nums[right]:
                right -= 1
            elif nums[mid] < nums[right]:
                right = mid
            else:
                left = mid + 1

        return nums[left]

Python：

class Solution2(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        left, right = 0, len(nums) - 1
        while left < right and nums[left] >= nums[right]:
            mid = left + (right - left) / 2
            
            if nums[mid] == nums[left]:
                left += 1
            elif nums[mid] < nums[left]:
                right = mid
            else:
                left = mid + 1

        return nums[left]　　

C++:

class Solution {
public:
    int findMin(vector<int> &nums) {
        if (nums.empty()) return 0;
        int left = 0, right = nums.size() - 1, res = nums[0];
        while (left < right - 1) {
            int mid = left + (right - left) / 2;
            if (nums[left] < nums[mid]) {
                res = min(res, nums[left]);
                left = mid + 1;
            } else if (nums[left] > nums[mid]) {
                res = min(res, nums[right]);
                right = mid;
            } else ++left;
        }
        res = min(res, nums[left]);
        res = min(res, nums[right]);
        return res;
    }
};

 

类似题目：

[LeetCode] 33. Search in Rotated Sorted Array 在旋转有序数组中搜索

[LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索 II

[LeetCode] 153. Find Minimum in Rotated Sorted Array 寻找旋转有序数组的最小值

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