LeetCode: 153. Find Minimum in Rotated Sorted Array

题目

原题地址: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/

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.

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.

题目大意是,给一个旋转过的数组(这个数组旋转前是个有序数组,旋转操作会把数组元素按循环往后移。 比如,旋转一次就是把元素往后移动一次,结果就是原来的最后一个元素后移一位变成了第一个元素,其他元素也都后移了一位), 找出这个数组中最小的那个元素。

解法

最简单的办法就是整个遍历一遍数组就可以找到最小的那个元素了。 不过这个办法没用利用题目中所说的数组旋转前是个有序数组的特性, 如果利用这个特性的话,可以减少遍历次数:

  • 二分查找,如果中间元素的值比右边尾部元素的值要小,说明旋转没有超过一半,最小值在左边,在左边元素中继续进行二分操作
  • 如果中间元素的值比右边尾部元素的值要大,说明旋转已经超过一半了,最小值在右边,在右边元素中继续进行二分查找
  • 二分到最后的那个元素就是最小的那个元素

这个思路的 Python 代码类似下面这样:

class Solution:
    def findMin(self, nums):
        if len(nums) <= 2:
            return min(nums)

        left = 0
        right = len(nums) - 1
        while left < right:
            mid = left + (right - left) // 2
            if nums[mid] < nums[right]:
                right = mid
            else:
                left = mid + 1

        return nums[right]

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