164. Maximum Gap

Given an unsorted array,find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,9],either
?            (3,6) or (6,9) has the maximum difference 3.

Example 2:

Input: [10]
Output: 0
Explanation: The array contains less than 2 elements,therefore return 0.

Note:

• You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
• Try to solve it in linear time/space.

  public class Solution {
      public int maximumGap(int[] nums) {
          if (nums.length < 2) return 0;
          bucketSort(nums);
  
          int maxDiff = Integer.MIN_VALUE;
          for (int i = 1; i < nums.length; ++i) {
              maxDiff = Math.max(maxDiff,nums[i] - nums[i - 1]);
          }
          return maxDiff;
      }
  
      private static void bucketSort(int[] nums) {
          if (nums.length < 2) return;
          int minValue = Integer.MAX_VALUE;
          int maxValue = Integer.MIN_VALUE;
  
          for (int i : nums) {
              minValue = Math.min(minValue,i);
              maxValue = Math.max(maxValue,i);
          }
          final int bucketSize = (maxValue - minValue) / nums.length + 1;
          final int bucketCount = (maxValue - minValue) / bucketSize + 1;
          final ArrayList<Integer>[] buckets = new ArrayList[bucketCount];
          for (int i = 0; i < buckets.length; ++i) {
              buckets[i] = new ArrayList<>();
          }
  
          for (int x : nums) {
              final int index = (x - minValue) / bucketSize;
              buckets[index].add(x);
          }
  
          for (final ArrayList<Integer> list : buckets) {
              Collections.sort(list);
          }
  
          int i = 0;
          for (final ArrayList<Integer> list : buckets) {
              for (int x : list) {
                  nums[i++] = x;
              }
          }
      }
  }

  桶排序：桶个数，桶容量，index

• final int bucketSize = (maxValue - minValue) / nums.length + 1;
  final int bucketCount = (maxValue - minValue) / bucketSize + 1;
  final int index = (x - minValue) / bucketSize;

  https://blog.csdn.net/afei__/article/details/82965834