153. 寻找旋转排序数组中的最小值

🟠   🔖  数组 二分查找  🔗 力扣 LeetCode

题目

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.

题目大意

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。

你可以假设数组中不存在重复元素。

解题思路

最直接的办法就是遍历一遍，找到最小值。但是还可以有更好的方法，可以用二分查找来降低算法的时间复杂度。

创建两个指针 left、right，分别指向数组首尾，然后计算出两个指针所指下标的中间值 mid，将 mid 与两个指针做比较。

• 如果 nums[mid] > nums[right]，则最小值不可能在 mid 左侧，一定在 mid 右侧，则将 left 移动到 mid + 1 位置，继续查找右侧区间。
• 如果 nums[mid] <= nums[right]，则最小值一定在 mid 左侧，或者 mid 位置，将 right 移动到 mid 位置上，继续查找左侧区间。

复杂度分析

• 时间复杂度：O(log n)
• 空间复杂度：O(1)

代码

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function (nums) {
	let left = 0,
		right = nums.length - 1;
	while (left < right) {
		let mid = Math.floor((left + right) / 2);
		if (nums[mid] > nums[right]) {
			left = mid + 1;
		} else {
			right = mid;
		}
	}
	return nums[left];
};

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