Questions:

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.

Thinking:

  • Method 1:
    • 先排序，再从头开始遍历找出max gap。
    • 并不满足题目要求的O（N），这种方法的复杂度是O(N + NlgN).
    • class Solution {
  public int maximumGap(int[] nums) {
  Arrays.sort(nums);
  int diff = 0;
  for(int i = 1; i < nums.length; i++)
      diff = Math.max(diff, nums[i] - nums[i - 1]);
  return diff;
  }
}

二刷

  1. 研究了桶排序的原理。

Imgur

class Solution {
    public int maximumGap(int[] nums) {
        if(nums.length <= 1) return 0;
        int max = Integer.MIN_VALUE, min = Integer.MAX_VALUE;
        for(int num : nums){
            max = Math.max(max, num);
            min = Math.min(min, num);
        }
        if(max == min) return 0;
        if(nums.length == 2) return max - min;
        int len = (int)Math.ceil((double)(max - min) / (nums.length - 1));
        int n = (max - min) / len;
        int[] minValues = new int[n + 1];
        int[] maxValues = new int[n + 1];
        for(int i = 0; i < n + 1; i++){
            minValues[i] = Integer.MAX_VALUE;
            maxValues[i] = Integer.MIN_VALUE;
        }
        for(int num : nums){
            int index = (num - min) / len;
            minValues[index] = Math.min(minValues[index], num);
            maxValues[index] = Math.max(maxValues[index], num);
        }
        int result = 0;
        int pre = -1;
        for(int i = 0; i < n + 1; i++){
            if(minValues[i] != Integer.MAX_VALUE){
                if(pre != -1)
                    result = Math.max(result, minValues[i] - pre);
                pre = maxValues[i];
            }
        }
        return result;
    }
}