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LeetCode 153 Find Minimum in Rotated Sorted Array

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LeetCode



Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times. [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.


Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.


Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.


Constraints:

n == nums.length 1 <= n <= 5000
-5000 <= nums[i] <= 5000 All the integers of nums are unique.
* nums is sorted and rotated between 1 and n times.

算法思路：

二分法

注意事项：

  1. 如果nums[start] > nums[mid]或nums[mid] > nums[end]都将会是min所在的区间。但是由于无位移数组的min在左边，所以优先判断后半区间nums[mid] > nums[end]，否则若用nums[start] > nums[mid] + else会忽略无位移情况。

Python代码：

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def findMin(self, nums: List[int]) -> int:
	start, end = 0, len(nums) - 1
	while start + 1 < end:
		mid = start + (end - start) // 2
		if nums[mid] > nums[end]:
			start = mid
		else:
			end = mid
	if nums[start] < nums[end]:
		return nums[start]
	else:
		return nums[end]

Java代码：

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public int findMin(int[] nums) {
	if(nums == null || nums.length == 0)
		return -1;
	
	int start = 0, end = nums.length - 1;
	while(start + 1 < end) {
		int mid = start + (end - start) / 2;
		if(nums[start] > nums[end]) {
			if (nums[mid] < nums[end])
				end = mid;
			else
				start = mid;
		} 
		else 
			return nums[start];
	}
	if(nums[start] < nums[end])
		return nums[start];
	else
		return nums[end];
}

算法分析：

时间复杂度为O(logn)，空间复杂度O(1)

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