LeetCode 164. Maximum Gap

原题链接在这里：https://leetcode.com/problems/maximum-gap/

题目：

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.

题解：

桶排序（bucket sort）

假设有N个元素A到B.

bucket（桶）的大小bucketLen = max(1,(B - A) / (N - 1)), 注意bucketLen为0的情况，则最多会有(B - A) / bucketLen + 1个桶

对于数组中的任意整数K, 很容易通过算式loc = (K - A) / bucketLen找出其桶的位置，然后维护每一个桶的最大值和最小值

由于同一个桶内的元素之间的差值至多为bucketLen - 1，因此最终答案不会从同一个桶中选择，否则整体的range达不到B-A.

对于每一个非空的桶p，找出下一个非空的桶q，则q.min - p.max可能就是备选答案。返回所有这些可能值中的最大值。

Note: corner case [1,1,1,1]若元素都相同，range就是0，16行算bucket时需要保持bucket的长度大于等于1.否则17行就有除以0的error.

同时此情况maxGap = 0. 所以maxGap的初始化是0而不是Integer.MIN_VALUE.

Time Complexity: O(n).

Space O(n).

AC Java:

public class Solution {
    public int maximumGap(int[] nums) {
        if(nums == null || nums.length < 2){
            return 0;
        }
        //Calculate the range
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        int len = nums.length;
        for(int i = 0; i<len; i++){
            max = Math.max(max,nums[i]);
            min = Math.min(min,nums[i]);
        }
        
        //bucket length and number
        int bucketLen = Math.max(1,(max-min)/(len-1)); //error
        int buckNum = (max-min)/bucketLen + 1;
        
        //calculate nums[i] position in buckets and maintain the maximun and minmum of each bucket
        int [] maxArr = new int[buckNum];
        int [] minArr = new int[buckNum];
        Arrays.fill(maxArr, Integer.MIN_VALUE);
        Arrays.fill(minArr, Integer.MAX_VALUE);
        for(int i = 0; i<len; i++){
            int loc = (nums[i]-min)/bucketLen;
            maxArr[loc] = Math.max(maxArr[loc],nums[i]);
            minArr[loc] = Math.min(minArr[loc],nums[i]);
        }
        //Calculate maximum gap
        int maxGap = 0;
        int preMax = maxArr[0];
        for(int i = 1; i<buckNum; i++){
            if(maxArr[i] == Integer.MIN_VALUE || minArr[i] == Integer.MAX_VALUE){
                continue;
            }
            maxGap = Math.max(maxGap, minArr[i]-preMax);
            preMax = maxArr[i];
        }
        return maxGap;
    }
}