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.

Credits:

Special thanks to @porker2008 for adding this problem and creating all test cases.

思路： 基本思想是利用桶排序。 

先找到max  和 min ， 然後平均 gap ＝  (int)Math.ceil((double)(max - min)/(num.length - 1));

因爲有可能好幾個數比較密集，在同一個 gap 裏面， 因此要用兩個桶來存放

    int[] bucketsMIN = new int[num.length - 1]; // store the min value in that bucket
    int[] bucketsMAX = new int[num.length - 1]; // store the max value in that bucket

將 int idx = (i - min) / gap; 

        bucketsMIN[idx] = Math.min(i, bucketsMIN[idx]);
        bucketsMAX[idx] = Math.max(i, bucketsMAX[idx]);

分別存入最大最小桶。

最後遍歷， 每次和 previous 比較， 

注意 初始化時候  previous 是 min，  loop 裏面 previous = bucketsMAX[i];

別忘記最後要 maxGap = Math.max(maxGap, max - previous);

public class Solution {
public int maximumGap(int[] num) {
    if (num == null || num.length < 2)
        return 0;
    // get the max and min value of the array
    int min = num[0];
    int max = num[0];
    for (int i:num) {
        min = Math.min(min, i);
        max = Math.max(max, i);
    }
    // the minimum possibale gap, ceiling of the integer division
    int gap = (int)Math.ceil((double)(max - min)/(num.length - 1));
    int[] bucketsMIN = new int[num.length - 1]; // store the min value in that bucket
    int[] bucketsMAX = new int[num.length - 1]; // store the max value in that bucket
    Arrays.fill(bucketsMIN, Integer.MAX_VALUE);
    Arrays.fill(bucketsMAX, Integer.MIN_VALUE);
    // put numbers into buckets
    for (int i:num) {
        if (i == min || i == max)
            continue;
        int idx = (i - min) / gap; // index of the right position in the buckets
        bucketsMIN[idx] = Math.min(i, bucketsMIN[idx]);
        bucketsMAX[idx] = Math.max(i, bucketsMAX[idx]);
    }
    // scan the buckets for the max gap
    int maxGap = Integer.MIN_VALUE;
    int previous = min;
    for (int i = 0; i < num.length - 1; i++) {
        if (bucketsMIN[i] == Integer.MAX_VALUE && bucketsMAX[i] == Integer.MIN_VALUE)
            // empty bucket
            continue;
        // min value minus the previous value is the current gap
        maxGap = Math.max(maxGap, bucketsMIN[i] - previous);
        // update previous bucket value
        previous = bucketsMAX[i];//----Probably more than 1 element in one position
    }
    maxGap = Math.max(maxGap, max - previous); // updata the final max value gap
    return maxGap;
}
}