154. Find Minimum in Rotated Sorted Array II

🔴   🔖  数组 二分查找  🔗 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 = [0,1,4,4,5,6,7] 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 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。
• 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。

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];
};

相关题目

相关题目

• 153. 寻找旋转排序数组中的最小值

- [153. 寻找旋转排序数组中的最小值](./0153.md)