153. Find Minimum in Rotated Sorted Array

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

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.

💻 Lời Giải

C++ 0153-find-minimum-in-rotated-sorted-array.cpp 
class Solution {
public:
    int findMin(vector<int>& nums) {
        int n = nums.size();
        int l = 0, r = n - 1;
        int i = 0;
        while (l <= r) {
            int m = l + (r - l) / 2;
            if (m > 0 and m + 1 < n and nums[m - 1] > nums[m] and nums[m + 1] > nums[m]) {
                return nums[m];
            }
            if (nums[m] > nums[r]) {
                l = m + 1;
            } else {
                i = m;
                r = m - 1;
            }
        }
        return nums[i];
    }
};