Find Minimum in Rotated Sorted Array I & II

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本文介绍了一种高效算法，用于在未知旋转点的排序数组中查找最小元素。包括两种情况：数组元素唯一和允许重复的情况，并提供了具体的Java实现。

Suppose a sorted array 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.

You may assume no duplicate exists in the array.

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public class Solution {
    public int findMin(int[] num) {
        if(num.length < 1)
            return 0;
        int left = 0;
        int right = num.length - 1;
        while(left < right){
            int mid = (left + right) /2;
            if(num[mid] > num[right]){
                left = mid + 1;//-------
            }else
                right = mid;
        }
        return num[left];
    }
}

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Suppose a sorted array 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.

public class Solution {
    public int findMin(int[] num) {
        if(num.length < 1)
            return 0;
        int left = 0;
        int right = num.length - 1;
        while(left < right && num[left] >= num[right]){
            int mid = (left + right)/2;
            if(num[mid] > num[right])
                left = mid + 1;
            else if(num[mid] < num[left])
                right = mid;
            else//num[left] == num[right] == num[mid]
                left++;
        }
        return num[left];
    }
}