(LeetCode)164. Maximum Gap

最新推荐文章于 2024-08-20 17:55:37 发布

原创最新推荐文章于 2024-08-20 17:55:37 发布 · 275 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

5 篇文章

订阅专栏

博客围绕未排序数组展开，要求找出其排序后相邻元素间的最大差值，若数组元素少于2个则返回0，还假定数组元素为非负整数且在32位有符号整数范围内，尝试以线性时间和空间复杂度求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
             (3,6) or (6,9) has the maximum difference 3.

Example 2:

Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.

Note:

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Try to solve it in linear time/space.

解法来自于评论区：

class Solution {
    public int maximumGap(int[] nums) {
        if(nums.length < 2) return 0;
        int min = nums[0]; 
        int max = nums[0];
        for(int i : nums)
        {
            min = Math.min(min, i);
            max = Math.max(max, i);
        }
        int gap = (int)Math.ceil((double)(max - min) / (nums.length - 1));
        
        int[] bucketsMIN = new int[nums.length - 1];
        int[] bucketsMAX = new int[nums.length - 1];
        
        Arrays.fill(bucketsMIN, Integer.MAX_VALUE);
        Arrays.fill(bucketsMAX, Integer.MIN_VALUE);
        
        for(int i : nums)
        {
            if(i == min || i == max) continue;
            else
            {
                int bucketid = (i - min) / gap;
                bucketsMIN[bucketid] = Math.min(bucketsMIN[bucketid], i);
                bucketsMAX[bucketid] = Math.max(bucketsMAX[bucketid], i);
            }
        }
        
        int maxgap = 0;
        int prev = min;
        for(int i = 0; i < nums.length - 1; i++)
        {
            if(bucketsMIN[i] == Integer.MAX_VALUE && bucketsMAX[i] == Integer.MIN_VALUE)
                continue;
            maxgap = Math.max(maxgap, bucketsMIN[i] - prev);
            prev = bucketsMAX[i];
        }
        maxgap = Math.max(maxgap, max - prev);
        return maxgap;
    }
}