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 [1,3,6,9], 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
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final int bucketSize = (maxValue - minValue) / nums.length + 1; final int bucketCount = (maxValue - minValue) / bucketSize + 1; final int index = (x - minValue) / bucketSize;