154 Find Minimum in Rotated Sorted Array II 寻找旋转排序数组中的最小值 II
Description:
Suppose an array sorted in ascending order 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.
Example:
Example 1:
Input: [1,3,5]
Output: 1
Example 2:
Input: [2,2,2,0,1]
Output: 0
Note:
This is a follow up problem to Find Minimum in Rotated Sorted Array.
Would allow duplicates affect the run-time complexity? How and why?
题目描述:
假设按照升序排序的数组在预先未知的某个点上进行了旋转。
( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
请找出其中最小的元素。
注意数组中可能存在重复的元素。
示例 :
示例 1:
输入: [1,3,5]
输出: 1
示例 2:
输入: [2,2,2,0,1]
输出: 0
说明:
这道题是 寻找旋转排序数组中的最小值 的延伸题目。
允许重复会影响算法的时间复杂度吗?会如何影响,为什么?
思路:
参考LeetCode #153 Find Minimum in Rotated Sorted Array 寻找旋转排序数组中的最小值
- 逐个搜索
时间复杂度O(n), 空间复杂度O(1) - 二分法
当nums[right] == nums[mid]时, --right
其他的思路跟上一题一样
时间复杂度O(n), 空间复杂度O(1), 平均时间复杂度O(lgn)
代码:
C++:
class Solution
{
public:
int findMin(vector<int>& nums)
{
int left = 0, right = nums.size() - 1;
while (left < right)
{
int mid = left + ((right - left) >> 1);
if (nums[mid] > nums[right]) left = mid + 1;
else if (nums[mid] < nums[right]) right = mid;
else if (nums[mid] == nums[right]) --right;
}
return nums[left];
}
};
Java:
class Solution {
public int findMin(int[] nums) {
int left = 0, right = nums.length - 1;
while (left < right) {
int mid = left + ((right - left) >>> 1);
if (nums[mid] > nums[right]) left = mid + 1;
else if (nums[mid] < nums[right]) right = mid;
else if (nums[mid] == nums[right]) --right;
}
return nums[left];
}
}
Python:
class Solution:
def findMin(self, nums: List[int]) -> int:
return min(nums)