LeetCode 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.

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题目大意：

给定一个未排序的数组。找出其排序后的序列中两个相邻元素之间的最大差值。

最好在线性时间、线性空间复杂度内完毕。

假设数组少于2个元素，返回0

能够如果数组中的全部元素均为非负整数，而且在32位带符号整数的范围以内。

解题思路：

基数排序（radix sort）/桶排序（bucket sort）

官方版（桶排序）：

如果有N个元素A到B。

那么最大差值不会大于ceiling[(B - A) / (N - 1)]

令bucket（桶）的大小len = ceiling[(B - A) / (N - 1)]，则最多会有(B - A) / len + 1个桶

对于数组中的随意整数K，非常easy通过算式loc = (K - A) / len找出其桶的位置，然后维护每个桶的最大值和最小值

因为同一个桶内的元素之间的差值至多为len - 1，因此终于答案不会从同一个桶中选择。

对于每个非空的桶p，找出下一个非空的桶q。则q.min - p.max可能就是备选答案。返回全部这些可能值中的最大值。

class Solution {
	public:
		int maximumGap(vector<int> &num) { 
			if (num.empty() || num.size() < 2)
				return 0;
			int Max = *max_element(num.begin(), num.end());
			int Min = *min_element(num.begin(), num.end());

			int gap = (int)ceil((double)(Max-Min)/(num.size() - 1));
			int bucketNum = (int) ceil((double)(Max-Min)/gap);
			vector<int> bucketsMin(bucketNum, INT_MAX);	
			vector<int> bucketsMax(bucketNum, INT_MIN);

			for (int i = 0; i < num.size(); i++) {
				if (num[i] == Max || num[i] == Min)
					continue;
				int idx = (num[i] - Min) / gap;
				bucketsMin[idx] = min(bucketsMin[idx], num[i]);
				bucketsMax[idx] = max(bucketsMax[idx], num[i]);
			}

			int ans = INT_MIN;
			int previous = Min;
			for (int i = 0; i < bucketNum; i++) {
				if (bucketsMin[i] == INT_MAX || bucketsMax[i] == INT_MIN) 
					continue;
				ans = max(ans, bucketsMin[i] - previous);
				previous = bucketsMax[i];
			}

			ans = max(ans, Max - previous);
			return ans;
		}
};