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