Find Minimum in Rotated Sorted Array II

最新推荐文章于 2022-05-26 16:47:57 发布

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最新推荐文章于 2022-05-26 16:47:57 发布

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分类专栏： Binary Search

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本文探讨在可能包含重复元素的已排序数组被未知旋转后，如何找到最小元素的算法。通过比较中间元素与结束元素，算法巧妙地在O(log n)时间内找到答案。

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Suppose an array sorted in ascending order 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.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

This is a follow up problem to Find Minimum in Rotated Sorted Array.
Would allow duplicates affect the run-time complexity? How and why?

Have you met this question in a real interview

思路：跟Find Minimum in Rotated Sorted Array I 相似，如果mid跟end相等，那就end --, 去掉end，也不会影响去掉最小的，因为有重复的；

class Solution {
    public int findMin(int[] nums) {
        if(nums == null || nums.length == 0) {
            return -1;
        }
        int start = 0; int end = nums.length - 1;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(nums[mid] == nums[end]) {
                end--;
            }else if(nums[mid] < nums[end]) {
                end = mid;
            } else {
                // nums[mid] > nums[end];
                start = mid;
            }
        }
        
        if(nums[start] < nums[end]) {
            return nums[start];
        }
        return nums[end];
    }
}