[LeetCode] Maximum Gap

This problem has a naive solution using sort and linear scan. The suggested solution uses the idea of bucket sort. The following is a C++ implementation of the suggested solution.

Suppose all the n elements in nums fall within [l, u], the maximum gap will not be smaller than gap = (u - l) / (n - 1). However, this gap may become 0 and so we take the maximum of it with 1 to guarantee that the gap used to create the buckets is meaningful.

Then there will be at most m = (u - l) / gap + 1 buckets. For each number num, it will fall in the k = (num - u) / gap bucket. After putting all elements of nums in the corresponding buckets, we can just scan the buckets to compute the maximum gap.

The maximum gap is only dependent on the maximum number of the current bucket and the minimum number of the next neighboring bucket (the bucket should not be empty). So we only store the minimum and the maximum of each bucket. Each bucket is initialized as {minimum = INT_MAX, maximum = INT_MIN} and then updated while updating the buckets.

Putting these together, we can have the following solution, barely a straight-forward implementation of the suggested solution.

复制代码
 1 class Solution {
 2 public:
 3     int maximumGap(vector<int>& nums) {
 4         int n = nums.size();
 5         if (n < 2) return 0;
 6         auto lu = minmax_element(nums.begin(), nums.end());
 7         int l = *lu.first, u = *lu.second;
 8         int gap = max((u - l) / (n - 1), 1);
 9         int m = (u - l) / gap + 1;
10         vector<vector<int>> buckets(m, {INT_MAX, INT_MIN});
11         for (int num : nums) {
12             int k = (num - l) / gap;
13             if (num < buckets[k][0]) buckets[k][0] = num;
14             if (num > buckets[k][1]) buckets[k][1] = num;
15         }
16         int i = 0, j;
17         gap = buckets[0][1] - buckets[0][0];
18         while (i < m) {
19             j = i + 1;
20             while (j < m && buckets[j][0] == INT_MAX && buckets[j][1] == INT_MIN)
21                 j++;
22             if (j == m) break;
23             gap = max(gap, buckets[j][0] - buckets[i][1]);
24             i = j;
25         }
26         return gap; 
27     }
28 };
复制代码

 

posted @   jianchao-li  阅读(277)  评论(0编辑  收藏  举报
编辑推荐:
· 深入理解 Mybatis 分库分表执行原理
· 如何打造一个高并发系统?
· .NET Core GC压缩(compact_phase)底层原理浅谈
· 现代计算机视觉入门之:什么是图片特征编码
· .NET 9 new features-C#13新的锁类型和语义
阅读排行:
· Sdcb Chats 技术博客:数据库 ID 选型的曲折之路 - 从 Guid 到自增 ID,再到
· 语音处理 开源项目 EchoSharp
· 《HelloGitHub》第 106 期
· Spring AI + Ollama 实现 deepseek-r1 的API服务和调用
· 使用 Dify + LLM 构建精确任务处理应用
点击右上角即可分享
微信分享提示