README_EN.md

153. Find Minimum in Rotated Sorted Array

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Description

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] 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 of unique elements, return the minimum element of this array.

 

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

 

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

Solutions

Python3

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

Java

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

C++

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

Go

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

JavaScript

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

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