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.

这题是桶排序的应用，其实不需要完全的排序。只需要计算出相邻桶之间的元素的差值。所以每个桶里面只需要维护两个元素即可，最大和最小的。

注意肯定有的桶是空桶，注意跳过。

桶的单位大小是(max-min)/nums.size()+1;

代码如下：

int maximumGap(vector<int>& nums) {
        if(nums.size()<2) return 0;
        int min=nums[0],max=nums[0];
        for(int i=1;i<nums.size();i++)
        {
            min=min>nums[i]?nums[i]:min;
            max=max<nums[i]?nums[i]:max;
        }
        //得到了min和max，可以分桶了
        int bucket_size=(max-min)/nums.size()+1;
        vector<vector<int> > bucket((max-min)/bucket_size+1);
        for(int i=0;i<nums.size();i++)
        {
            int index=(nums[i]-min)/bucket_size;
            if(bucket[index].empty())
            {
                bucket[index].push_back(nums[i]);
                bucket[index].push_back(nums[i]);
            }else{
                bucket[index][0]= bucket[index][0]>nums[i]?nums[i]:bucket[index][0];
                bucket[index][1]= bucket[index][1]<nums[i]?nums[i]:bucket[index][1];
            }
        }
        int prev=0;
        int maxgap=INT_MIN;
        for(int i=1;i<bucket.size();i++)
        {
            if(bucket[i].empty()) continue;
            int gap=bucket[i][0]-bucket[prev][1];
            prev=i;
            maxgap=maxgap<gap?gap:maxgap;
        }
        return maxgap;
    }

leetcode的solution说明：

Suppose there are N elements and they range from A to B.

Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]

Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket

for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.

Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.

For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.