Maximum Gap (ARRAY - SORT)

QUESTION

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.

1ST TRY

桶排序

class Solution {

public:

    int maximumGap(vector<int> &num) {

        if(num.empty() || num.size() < 2)

            return 0;

        int maxNum = *max_element(num.begin(), num.end());

        int minNum = *min_element(num.begin(), num.end());

        

        //bucket gap: 假设每个数一个桶，两个桶之间的平均差值

        int gap = ceil((double)(maxNum - minNum)/(num.size()-1));

        //number of buckets

        int bucketNum = (maxNum-minNum)/gap+1;

        //declare buckets

        vector<int> bucketsMin(bucketNum, INT_MAX);

        vector<int> bucketsMax(bucketNum, INT_MIN);

        //put into buckets

        for(int i = 0; i < num.size(); i ++)

        {

            int buckInd = (num[i]-minNum)/gap; //匹配到bucket

            bucketsMin[buckInd] = min(bucketsMin[buckInd], num[i]);

            bucketsMax[buckInd] = max(bucketsMax[buckInd], num[i]);

        }

        

        //i_th gap is minvalue in i+1_th bucket minus maxvalue in i_th bucket 

        int maxGap = INT_MIN;

        int previous = minNum;

        for(int i = 0; i < bucketNum; i ++)

        {

            if(bucketsMin[i] == INT_MAX && bucketsMax[i] == INT_MIN)

                continue;   //empty

            maxGap = max(maxGap, bucketsMin[i]-previous);

            previous = bucketsMax[i];

        }

        return maxGap;

    }

};

Result: Accepted