leetcode | Maximum Gap

Maximum Gap: https://leetcode.com/problems/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.

e.g. [1, 3, 6, 9] return 3.

解析：

题目要求，找出排序后的相邻元素间最大差值。那么首先不得不做的就是排序，而常规的比较排序算法，不能突破O(nlgn)O(nlgn)，要满足题目要求的线性复杂度，只能选用线性排序算法。

线性排序算法：计数排序，基数排序，桶排序，点击查看线性排序算法

由于待排序数的范围可能较大，如果使用计数排序或桶排序可能需要较大的空间。

本题采用基数排序 ： 时间复杂度O(kn) 空间复杂度 O(n)

class Solution {
public:
int maximumGap(vector<int>& nums) {
int size = nums.size();
if (size < 2)
return 0;
radixSort(nums);
int *gap = new int[size - 1];
for (int i = 0; i < size - 1; i++)
gap[i] = nums[i + 1] - nums[i];
int maxGap = gap[0];
for (int i = 1; i < size - 1; i++) {
if (gap[i] > maxGap)
maxGap = gap[i];
}
delete []gap;
return maxGap;
}
private:
void radixSort(vector<int>& nums) {
int size = nums.size();
int d = getMaxBits(nums); // 得到最大数的位数
int *count = new int[10]; // 10个桶 0~9
int* tmp = new int[size];
int radix = 1;
for (int i = 0; i < d; i++) {
for (int j = 0; j < 10; j++)
count[j] = 0;
for (int j = 0; j < size; j++) {
int k = nums[j] / radix % 10;
count[k]++;
}
for (int j = 1; j < 10; j++)
count[j] += count[j - 1];
for (int j = size - 1; j >= 0; j--) {
int k = nums[j] / radix % 10;
tmp[count[k] - 1] = nums[j];
count[k]--;
}
for (int j = 0; j < size; j++)
nums[j] = tmp[j];
radix *= 10;
}
delete []count;
delete []tmp;

}
// 计算最高位
int getMaxBits(vector<int>& nums) {
int size = nums.size();
int max_data = nums[0];
for (int i = 1; i < size; i++) {
if (nums[i] > max_data)
max_data = nums[i];
}
int d = 1;
int radix = 10;
while (max_data >= radix) {
d++;
radix *= 10;
}
return d;
}
};