README_EN.md

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.

 

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 is the same as Find Minimum in Rotated Sorted Array but with duplicates. Would allow duplicates affect the run-time complexity? How and why?

Solutions

Python3

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

Java

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

C++

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

JavaScript

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

Go

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

...