# LeetCode 笔记28 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.

public int maximumGap(int[] num) {
if (num == null || num.length <= 1) {
return 0;
}

for (int d = 0; d < 32; d++) {
int[] count = new int[3];
int[] aux = new int[num.length];
for (int i = 0; i < num.length; i++) {
count[((num[i] >> d) & 1) + 1]++;
}

for (int i = 1; i < 2; i++) {
count[i] += count[i - 1];
}

for (int i = 0; i < num.length; i++) {
aux[count[((num[i] >> d) & 1)]++] = num[i];
}

for (int i = 0; i < num.length; i++) {
num[i] = aux[i];
}
}
int maxGap = 0;
for (int i = 1; i < num.length; i++) {
if (num[i] - num[i - 1] > maxGap) {
maxGap = num[i] - num[i - 1];
}
}
return maxGap;
}

count数组的意义具体可以参考上面提到的书关于String Sort的第一部分。

public int maximumGap2(int[] num) {
if (num == null || num.length <= 1) {
return 0;
}
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
for (int i = 0; i < num.length; i++) {
if (num[i] > max) {
max = num[i];
}
if (num[i] < min) {
min = num[i];
}
}
int len = (max - min) / (num.length - 1);
if (len == 0) {
len = 1;
}
int numOfBucket = (max - min) / len + 1;
Bucket[] buckets = new Bucket[numOfBucket];
for (int i = 0; i < num.length; i++) {
int idx = (num[i] - min) / len;
if (buckets[idx] == null) {
buckets[idx] = new Bucket();
}
if (num[i] > buckets[idx].max) {
buckets[idx].max = num[i];
}
if (num[i] < buckets[idx].min) {
buckets[idx].min = num[i];
}
}
int maxGap = 0;
max = -1;
for (int i = 0; i < buckets.length; i++) {
if (buckets[i] != null) {
if (max == -1) {
//pass
} else {
maxGap = Math.max(buckets[i].min - max, maxGap);
}
max = buckets[i].max;
}
}
return maxGap;
}

However，我发现桶排序做出来好像比基数排序慢也。

posted on 2015-03-11 22:19  lichen782  阅读(...)  评论(...编辑  收藏