LeetCode(164) Maximum Gap

題目如下：

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Try to solve it in linear time/space.
Return 0 if the array contains less than 2 elements.
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

分析如下：

考慮了很多方法，估計要用到bucket sort或者radix sort。後來放狗一搜發現果然要用到bucket sort.

基本思路是這樣的，把n個數放到x個桶中。我們可以構造一個合適的x，並且通過下面的證明發現，max gap只可能出現在兩個桶之間，不能出現在一個桶的內部。

下面是證明，直接抄官方的解答：

Suppose there are N elements and they range from A to B.

Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]

Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket

for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.

Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.

For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.

My Code：

//48msstruct Bucket {    int local_min;    int local_max;    int local_volumn;    Bucket():local_volumn(0){};};class Solution {public:    int maximumGap(vector<int> &num) {        if (num.size() < 2) return 0;        if (num.size() == 2) return num[0] - num[1] > 0 ? num[0] - num[1] : num[1] - num[0];        int min = INT_MAX;        int max = INT_MIN;        int max_gap = 0;        for (int i = 0; i < num.size(); ++i) {            if (num[i] < min) min = num[i];            if (num[i] > max) max = num[i];        }        int bucket_length = (max - min) / (num.size() - 1) + 1; //maximum gap >= ceiling[(B - A) / (N - 1)]//        int bucket_length = (max - min) / num.size() + 1; //居然也能AC......        int bucket_count = (max - min) / bucket_length + 1;        vector<Bucket> bucket(bucket_count, Bucket());        for (int i = 0; i < num.size(); ++i) {            int index = (num[i] - min) / bucket_length;            if (bucket[index].local_volumn == 0) //NOTE1: 如果桶內最多隻有一個元素，那麼max 和 min都賦值為這個元素。                bucket[index].local_max = bucket[index].local_min = num[i];            else if (num[i] > bucket[index].local_max)                bucket[index].local_max = num[i];            else if (num[i] < bucket[index].local_min)                bucket[index].local_min = num[i];            bucket[index].local_volumn++;        }        int previous = 0; //NOTE2: 用previous來跳過空桶,第0個桶必然不是空桶，所以previous可以從0開始取值。        for (int i = 1; i < bucket.size(); ++i) {            if (bucket[i].local_volumn == 0) continue;            if (bucket[i].local_min  - bucket[previous].local_max > max_gap)                max_gap = bucket[i].local_min  - bucket[previous].local_max;            previous = i;        }        return max_gap;    }};

小結擴充：

1 排序的基本思想可以擴充出很多內容。

2 經過測試發現

bucket_length = (max - min) / (num.size() - 1) + 1;

如果寫為

bucket_length = (max - min) / num.size() + 1;

也可以AC。

但是從意義上來說，第1種寫法我覺得才是正確的。第2種暫時還沒有想到反例去推翻，但是Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]， 所以從意義上來說我覺得不正確。

3 題目中說32-bit，我覺得強烈暗示了可以用bit方式進行radix sort，這個思路還沒實現。