[Leetcode] Maximum Gap

Maximum Gap

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.

Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.

桶排序，由于num中的数字肯定在[min,max]区间内，所以根据抽屉原理，假设num中有n个数字，则最大的gap必然要大于dis=(max-min)/(n-1)，所以我们可以把num所在的范围分成等间隔的区间，相邻区间内的元素之间的最大差值，即为要寻找的gap。

class Solution {
public:
    int maximumGap(vector<int> &num) {
        int n = num.size();
        if (n < 2) return 0;
        if (n == 2) return abs(num[0] - num[1]);
        int imin = num[0], imax = num[0];
        for (int i = 0; i < n; ++i) {
            imin = min(imin, num[i]);
            imax = max(imax, num[i]);
        }
        int dist = (imax-imin) / n + 1;
        vector<vector<int> > bucket((imax-imin)/dist + 1);
        int idx;
        for (int i = 0; i < n; ++i) {
            idx = (num[i] - imin) / dist;
            if (bucket[idx].empty()) {
                bucket[idx].push_back(num[i]);
                bucket[idx].push_back(num[i]);
            } else {
                bucket[idx][0] = min(bucket[idx][0], num[i]);
                bucket[idx][1] = max(bucket[idx][1], num[i]);
            }
        }
        int gap = 0, pre = 0, tmp;
        for (int i = 1; i < bucket.size(); ++i) {
            if (bucket[i].empty()) continue;
            tmp = bucket[i][0] - bucket[pre][1];
            gap = max(gap, tmp);
            pre = i;
        }
        return gap;
    }
};