Question

153. Find Minimum in Rotated Sorted Array

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 = [0,1,2,4,5,6,7] 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_.

You must write an algorithm that runs in O(log n) time.

 

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

Solution 1

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

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

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

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

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

impl Solution {
  pub fn find_min(nums: Vec<i32>) -> i32 {
    let mut left = 0;
    let mut right = nums.len() - 1;
    while left < right {
      let mid = left + (right - left) / 2;
      if nums[mid] > nums[right] {
        left = mid + 1;
      } else {
        right = mid;
      }
    }
    nums[left]
  }
}

/**
 * @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];
};