Question

154. Find Minimum in Rotated Sorted Array II

中文文档

Description

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?

 

Solutions

Solution 1

class Solution:
  def findMin(self, nums: List[int]) -> int:
    left, right = 0, len(nums) - 1
    while left < right:
      mid = (left + right) >> 1
      if nums[mid] > nums[right]:
        left = mid + 1
      elif nums[mid] < nums[right]:
        right = mid
      else:
        right -= 1
    return nums[left]

class Solution {
  public int findMin(int[] nums) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (nums[mid] > nums[right]) {
        left = mid + 1;
      } else if (nums[mid] < nums[right]) {
        right = mid;
      } else {
        --right;
      }
    }
    return nums[left];
  }
}

class Solution {
public:
  int findMin(vector<int>& nums) {
    int left = 0, right = nums.size() - 1;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (nums[mid] > nums[right])
        left = mid + 1;
      else if (nums[mid] < nums[right])
        right = mid;
      else
        --right;
    }
    return nums[left];
  }
};

func findMin(nums []int) int {
  left, right := 0, len(nums)-1
  for left < right {
    mid := (left + right) >> 1
    if nums[mid] > nums[right] {
      left = mid + 1
    } else if nums[mid] < nums[right] {
      right = mid
    } else {
      right--
    }
  }
  return nums[left]
}

function findMin(nums: number[]): number {
  let left = 0,
    right = nums.length - 1;
  while (left < right) {
    const mid = (left + right) >> 1;
    if (nums[mid] > nums[right]) {
      left = mid + 1;
    } else if (nums[mid] < nums[right]) {
      right = mid;
    } else {
      --right;
    }
  }
  return nums[left];
}

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