Leetcode 164 Maximum Gap 桶排序好题

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

给定数组，找出数组排序后两个相邻数字的差的最大值是多少。

难点在于要求线性时间和空间复杂度完成。

不加这个要求，直接快排就可以了，加了这个要求怎么办呢？我也想了很久......

想了好久，看了题解，发现还是思维僵化，根本就没有想到桶排序。

引用官方题解中的话

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

当所有元素均匀分布的时候，相邻数字的差的最大值最小，因此可以构造桶大小的为[(B - A) / (N - 1)]，

同一个桶内的元素的差一定小于桶的大小，这样就不需要管桶内的大小关系了，只需要关注每个桶的最小值和前一个非空桶的最大值之差就可以了。

class Solution {public:    int maximumGap(vector<int>& nums) {        if(nums.size()<2) return 0;        int minn = INT_MAX;        int maxx = INT_MIN;        for(int i = 0; i<nums.size(); i++)        {            minn = min(minn, nums[i]);            maxx = max(maxx, nums[i]);        }        double len = (maxx - minn)*1.0 / (nums.size() - 1);        if(len == 0) return 0;//防止所有元素一样大        int cnt = floor((maxx - minn) / len + 1);        vector<int> minb(cnt, INT_MAX);        vector<int> maxb(cnt, INT_MIN);        for(int i = 0;i<nums.size();i++) //元素分入桶        {            int id = floor((nums[i] - minn) / len);            minb[id] = min(minb[id], nums[i]);            maxb[id] = max(maxb[id], nums[i]);        }        int res = 0, premax = maxb[0];//第一个桶包含最小值，因此一定不为空        for(int i = 1;i<cnt; i++)        {            if(minb[i] != INT_MAX)            {                res = max(res, minb[i] - premax);                premax = maxb[i];            }        }        return res;    }};