Maximum Gap

Question:
http://www.lintcode.com/en/problem/maximum-gap/
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.
Sort is easy but will cost O(nlogn) time. Try to solve it in linear time and space.
Answer:

class Solution {  
public:  
    int maximumGap(vector<int> &num) {  
       if (num.size() < 2) return 0;  
        int maxNum = num[0];  
        int minNum = num[0];  
        for (int x : num) {  
            maxNum = max(maxNum, x);  
            minNum = min(minNum, x);  
        }  
        int len = (maxNum - minNum) / num.size() + 1;  
        vector<vector<int>> buckets((maxNum - minNum) / len + 1);  
        for (int x : num) {  
            int i = (x - minNum) / len;  
            if (buckets[i].empty()) {  
                buckets[i].reserve(2);  
                buckets[i].push_back(x);  
                buckets[i].push_back(x);  
            } else {  
                if (x < buckets[i][0]) buckets[i][0] = x;  
                if (x > buckets[i][1]) buckets[i][1] = x;  
            }  
        }  
        int gap = 0;  
        int prev = 0;  
        for (int i = 1; i < buckets.size(); i++) {  
            if (buckets[i].empty()) continue;  
            gap = max(gap, buckets[i][0] - buckets[prev][1]);  
            prev = i;  
        }  
        return gap;  
    }  
};