leetcode 164. Maximum Gap 最大间隔 + 一个很好的桶排序示范

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.

这道题就是在一个无序的数组中寻找最大的相邻元素的Gap，最直接的方法就是先排序。然后遍历即可。

但是这不是本体的要求，我想了很久没想出来，看了网上的答案，才明白这道题考查的是桶排序，根据数组的最大值最小值的落差以及元素的数量，分配一定数量的桶，然后比较相邻桶的最小值和最大值（上一个桶的最大值和下一个桶的最小值一定是相邻的）就可以得出最大的相邻元素的Gap。

注意这道题是桶排序的一个很好的示范，很值得学习。

代码如下：

import java.util.Arrays;/* * http://blog.csdn.net/tofu_jelly/article/details/43339305 *  * 借鉴桶排序 *  * */public class Solution {    public int maximumGap(int[] nums)     {        if(nums==null || nums.length<=1)            return 0;        int maxNum=Integer.MIN_VALUE,minNum=Integer.MAX_VALUE;        for(int i=0;i<nums.length;i++)        {            maxNum=Math.max(maxNum, nums[i]);            minNum=Math.min(minNum, nums[i]);        }        if(nums.length==2)            return maxNum-minNum;        //元素之间的最大间隔，注意向上取整        int averageGap = (int)Math.ceil((double)(maxNum-minNum)/(nums.length-1));        //这个是处理元素都相等的情况        if(averageGap==0)            return 0;        int numOfBuckets = (maxNum-minNum)/averageGap +1;        int []minBuck=new int[numOfBuckets];        Arrays.fill(minBuck, Integer.MAX_VALUE);        int []maxBuck=new int[numOfBuckets];        Arrays.fill(maxBuck, Integer.MIN_VALUE);        //minBuck和maxBuch放的是对应的最大值和最小值        for(int i=0;i<nums.length;i++)        {            int index=(nums[i]-minNum) / averageGap;            minBuck[index] = Math.min(minBuck[index], nums[i]);            maxBuck[index] = Math.max(maxBuck[index], nums[i]);        }        int maxGap = Integer.MIN_VALUE;        //上一个桶的最大值和下一个桶的最小值肯定是相邻的，所以才可能存在最大的gap        int pre=maxBuck[0];        for(int i=0;i<numOfBuckets;i++)        {            //桶为空的情况            if(minBuck[i]==Integer.MAX_VALUE && maxBuck[i]==Integer.MIN_VALUE)                continue;            else            {                maxGap=Math.max(maxGap,minBuck[i]-pre);                pre=maxBuck[i];            }           }        return maxGap;    }    /*     * 使用排序做的答案     * */    public int maximumGapUseSort(int[] nums)     {        if(nums==null || nums.length<=1)            return 0;        Arrays.sort(nums);        int maxGap=0;        for(int i=1;i<nums.length;i++)            maxGap=Math.max(nums[i]-nums[i-1], maxGap);        return maxGap;    }}

下面是C++的做法，这道题是桶排序的一个很好的应用

代码如下：

#include <iostream>#include <algorithm>#include <climits>#include <vector>using namespace std;//  164. Maximum Gapclass Solution {public:    int maximumGap(vector<int>& nums)    {        if (nums.size() <= 1)            return 0;        int minNum = nums[0], maxNum = nums[0];        for (int a : nums)        {            minNum = min(minNum, a);            maxNum = max(maxNum, a);        }        if (nums.size() == 2)            return maxNum - minNum;        int averageGap = (int)ceil((double)(maxNum - minNum)/(nums.size()-1));        if (averageGap == 0)            return 0;        int numBucks = (maxNum - minNum) / averageGap + 1;        vector<int> minBuck(numBucks, numeric_limits<int>::max());        vector<int> maxBuck(numBucks, numeric_limits<int>::min());        for (int a : nums)        {            int index = (a - minNum) / averageGap;            minBuck[index] = min(minBuck[index], a);            maxBuck[index] = max(maxBuck[index], a);        }        int maxGap = numeric_limits<int>::min();        int pre = maxBuck[0];        for (int i = 0; i < numBucks; i++)        {            if (minBuck[i] == numeric_limits<int>::max()                && maxBuck[i] == numeric_limits<int>::min())                continue;            else            {                maxGap = max(maxGap,minBuck[i] - pre);                pre = maxBuck[i];            }        }        return maxGap;    }};