README_EN.md

154. Find Minimum in Rotated Sorted Array II

中文文档

Description

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e.,  [0,1,2,4,5,6,7] might become  [4,5,6,7,0,1,2]).

Find the minimum element.

The array may contain duplicates.

Example 1:

Input: [1,3,5]

Output: 1

Example 2:

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

Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• 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];
};

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