[leetcode] Maximum Gap

From : https://leetcode.com/problems/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.

思路 ： 

用桶排序原理，将n个数放入n个桶中，记录每个桶中的最大和最小值。

如果n个数分别在n个桶中，那么后一桶的最大（同时也是最小）与前一桶的最小（同时也是最大）之差的最大就是maxGap； 否则，如果n个数不全在n个桶中，那么必然有空桶，空桶后一个的最小和前一个的最大的差必然比桶的长度达，那么后桶最小和前桶最大之差的最大仍然是maxGap。

综上所述，取 max{buckets[0].max-buckets[0].min,  buckets[n-1].max-buckets[n-1].min,  max {buckets[i].min-buckets.max}}， 即buckets[0].max-buckets[0].min   ，  buckets[n-1].max-buckets[n-1].min 和 max {buckets[i].min-buckets.max}的最大值。

public class Solution {        class Bucket {        // if min < 1， the bucket is empty        int min = -1;        int max;    }        public int maximumGap(int[] nums) {        if(nums.length < 2) {            return 0;        }        int len = nums.length;        int gmax = nums[0], gmin = gmax;        for(int i=1; i<len; ++i) {            int c = nums[i];            if(c < gmin) {                gmin = c;            } else if(c > gmax) {                gmax = c;            }        }                int size = (int) Math.ceil( (double)(gmax - gmin + 1) / (double)len );        Bucket[] buckets = new Bucket[len]; // len buckets        for(int i=0; i<len; ++i) {        buckets[i] = new Bucket();        }        for(int num : nums) {            Bucket bucket = buckets[(num-gmin)/size];            if(bucket.min < 0) {                // empty bucket                bucket.min = bucket.max = num;            } else {                if(num < bucket.min) {                    bucket.min = num;                } else if(num > bucket.max) {                    bucket.max = num;                }            }        }                int min = buckets[0].min, max=min, maxGap = buckets[0].max-min;        for(Bucket b : buckets) {            if(b.min < 0) {                continue;            }            min = b.min;            if(min - max > maxGap) {                maxGap = min-max;            }            max = b.max;        }                return gmax-min > maxGap ? gmax-min : maxGap;    }}