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154. 寻找旋转排序数组中的最小值 II


154. 寻找旋转排序数组中的最小值 IIopen in new window

🔴   🔖  数组 二分查找  🔗 力扣open in new window LeetCodeopen in new window

题目

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

  • [4,5,6,7,0,1,4] if it was rotated 4 times.
  • [0,1,4,4,5,6,7] 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 that may contain duplicates , return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

Example 1:

Input: nums = [1,3,5]

Output: 1

Example 2:

Input: nums = [2,2,2,0,1]

Output: 0

Constraints:

  • n == nums.length
  • 1 <= n <= 5000
  • -5000 <= nums[i] <= 5000
  • nums is sorted and rotated between 1 and n times.

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

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。

注意数组中可能存在重复的元素。

解题思路

这一题是第 153 题的加强版,增加了重复元素的条件,做法没有变,还是用二分搜索,只不过在相等元素上多增加一个判断即可。

创建两个指针 leftright,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。
  • 如果 nums[mid] == nums[right],无法判断在 mid 的哪一侧,可以采用 right = right - 1 逐步缩小区域。

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((right + left) / 2);
		if (nums[mid] > nums[right]) {
			left = mid + 1;
		} else if (nums[mid] < nums[right]) {
			right = mid;
		} else {
			right--;
		}
	}
	return nums[left];
};

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153寻找旋转排序数组中的最小值open in new window[✓]数组 二分查找