# 164. Maximum Gap

**<font color=red>难度:Hard<font>**

## 刷题内容
> 原题连接

* https://leetcode.com/problems/maximum-gap/

> 内容描述

```
Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
             (3,6) or (6,9) has the maximum difference 3.
Example 2:

Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.
Note:

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Try to solve it in linear time/space.
```

> 思路1
******- 时间复杂度: O(NlgN)******- 空间复杂度: O(1)******

用普通的方法的去做其实不难，先对数组进行排序，再遍历数组，依次计算 nums[i] - nums[i - 1] 求出最大值即可

```cpp
class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)
            return 0;
        sort(nums.begin(),nums.end());
        int ans = INT_MIN;
        for(int i = 1;i < nums.size();++i)
            ans = max(ans,nums[i] - nums[i - 1]);
        return ans;
    }
};
```

> 思路2
******- 时间复杂度: O(N)******- 空间复杂度: O(N)******

但是题目中的note已经提示了在线性的时间和空间复杂度完成。由于这题内的数字的大小都是在32bits以内的，那么我们可以用桶排序或者基数排序，具体的算法和复杂度的推导可以参考《算法导论》。这里我们采用了基数排序。
```cpp
class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)
            return 0;
        int res[nums.size()];
        int t[nums.size()][10];
        for(int i = 0;i < nums.size();++i)
        {
            res[i] = i;
            for(int j = 0;j < 10;++j)
                t[i][j] = 0;
            int temp = nums[i],j = 9;
            while(temp)
            {
                t[i][j--] = temp % 10;
                temp /= 10;
            }
        }
        for(int i = 9;i >= 0;--i)
        {
            int bucket[nums.size()],count1[10];
            memset(count1,0,sizeof(count1));
            for(int j = 0;j < nums.size();++j)
                count1[t[res[j]][i]]++;
            for(int j = 1;j < 10;++j)
                count1[j] += count1[j - 1];
            for(int j = nums.size() - 1;j >= 0;--j)
                bucket[--count1[t[res[j]][i]]] = res[j];
            for(int j = 0;j < nums.size();++j)
                res[j] = bucket[j];
        }
        int ans INT_MIN;
        for(int i = 1;i < nums.size();++i)
            ans = max(ans,nums[res[i]] - nums[res[i - 1]]);
        return ans;
    }
};
```