相邻两数最大差值

有一个×××数组A，请设计一个复杂度为O(n)的算法，算出排序后相邻两数的最大差值。
给定一个int数组A和A的大小n，请返回最大的差值。保证数组元素多于1个。

测试样例：

[1,2,5,4,6],5
返回：2
我的提交

-- coding:utf-8 --

class Gap:
def maxGap(self, A, n):

write code here

    minValue = min(A)
    maxValue = max(A)
    bucketNum = (maxValue - minValue ) // 2 + 1
    buckets = [[] for _ in range(bucketNum)]

    for num in A:
        buckets[(num - minValue) // 2].append(num)

    maxdiff = 0
    lastMax = max(buckets[0])
    nowMin = 0
    for i in range(1, bucketNum):
        if buckets[i] == []:
            continue
        nowMin = min(buckets[i])
        maxdiff = max(maxdiff, nowMin - lastMax)
        lastMax = max(buckets[i])
    return maxdiff

if name == 'main':
g = Gap()
print(g.maxGap([1,2,5,4,6],5))
参考答案
python

-- coding:utf-8 --

class Gap:
def maxGap(self, A, n):

write code here

    maxVal,minVal = max(A),min(A)
    hashTable = [[maxVal,minVal,0] for i in xrange(0,n + 1)]
    for x in A:
        ind = (x - minVal) * n // (maxVal - minVal)
        hashTable[ind][2] += 1
        hashTable[ind][0] = min(hashTable[ind][0],x)
        hashTable[ind][1] = max(hashTable[ind][1],x)
    diffMax = 0
    preMax = maxVal
    for x in hashTable:
        if x[2] == 0:
            continue
        diffMax = max(x[0] - preMax,diffMax)
        preMax = x[1]
    return diffMax

java
import java.util.*;

public class Gap {
public int maxGap(int[] nums,int N) {
if (nums == null || nums.length < 2) {
return 0;
}
int len = nums.length;
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (int i = 0; i < len; i++) {
min = Math.min(min, nums[i]);
max = Math.max(max, nums[i]);
}
if (min == max) {
return 0;
}
boolean[] hasNum = new boolean[len + 1];
int[] maxs = new int[len + 1];
int[] mins = new int[len + 1];
int bid = 0;
for (int i = 0; i < len; i++) {
bid = bucket(nums[i], len, min, max); // 算出桶号
mins[bid] = hasNum[bid] ? Math.min(mins[bid], nums[i]) : nums[i];
maxs[bid] = hasNum[bid] ? Math.max(maxs[bid], nums[i]) : nums[i];
hasNum[bid] = true;
}
int res = 0;
int lastMax = 0;
int i = 0;
while (i <= len) {
if (hasNum[i++]) { // 找到第一个不空的桶
lastMax = maxs[i - 1];
break;
}
}
for (; i <= len; i++) {
if (hasNum[i]) {
res = Math.max(res, mins[i] - lastMax);
lastMax = maxs[i];
}
}
return res;
}

// 使用long类型是为了防止相乘时溢出
public int bucket(long num, long len, long min, long max) {
    return (int) ((num - min) * len / (max - min));
}

}