Likou practice question——Find Minimum in Rotated Sorted Array

Find Minimum in Rotated Sorted Array

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.

answer

From the question, we can know that the given array is transformed from an increasing array through multiple rounds of rotate . Each round of rotate will[a[0], a[1], a[2], ..., a[n-1]]转化成[a[n-1], a[0], a[1], a[2], ..., a[n-2]]。

因此，经过多轮rotate后的数组nums一定有以下关系：nums[0]~nums[i]递增；nums[i+1]~nums[n-1]递增。因此想要找到数组nums的最小值，必须要找到下标i所在位置。题目要求时间复杂度为O(logn)，我们可以利用二分法进行搜索。

The search plan is as follows:

while(l<=r){
    mid=l+(r-l)/2;
    if(nums[mid]<nums[mid-1]&&nums[mid]<nums[mid+1]){
        return nums[mid];
    }else{
        if(nums[mid]>nums[n-1]){
             l=mid+1;
        }else{
            r=mid-1;
        }
    }
}

Complete code

class Solution {
public:
    int findMin(vector<int>& nums) {
        // nums[i:j] ascending ; nums[j+1:n] ascending
        int n=nums.size();
        if(n==1){
            return nums[0];
        }else if(n<3){
            return min(nums[0],nums[1]);
        }
        int l=1,r=n-2,mid,ans;
        while(l<=r){
            mid=l+(r-l)/2;
            if(nums[mid]<nums[mid-1]&&nums[mid]<nums[mid+1]){
                return nums[mid];
            }else{
                if(nums[mid]>nums[n-1]){
                    l=mid+1;
                }else{
                    r=mid-1;
                }
            }
        }
        return min(nums[0],nums[n-1]);
    }
};