【刷题-LeetCode】164 Maximum Gap

164. Maximum Gap

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.

解法1 (O(nlog n))排序

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

解法2 (O(n))排序算法：桶排序、计数、基数。max-min可能会比较大，计数排序可能需要很大的空间，实现后发现会超时。

解法2.1 基数排序

基数排序是按照低位先排序，然后收集；再按照高位排序，然后再收集；依次类推，直到最高位。有时候有些属性是有优先级顺序的，先按低优先级排序，再按高优先级排序。最后的次序就是高优先级高的在前，高优先级相同的低优先级高的在前。

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)return 0;
        radix_sort(nums);
        int res = 0;
        for(int i = 1; i < nums.size(); ++i)res = max(res, nums[i]-nums[i-1]);
        return res;
    }
    void radix_sort(vector<int>& nums){
        int maxVal = *max_element(nums.begin(), nums.end());
        int exp = 1, radix = 10;
        vector<int>aux(nums.size());
        while(maxVal / exp > 0){
            vector<int>cnt(radix, 0);
            for(int i = 0; i < nums.size(); ++i){
                int idx = (nums[i] / exp) % radix;
                cnt[idx]++;
            }
            for(int i = 1; i < cnt.size(); ++i)cnt[i] += cnt[i-1];
            for(int i = nums.size() - 1; i >= 0; --i){
                aux[--cnt[(nums[i] / exp) % 10]] = nums[i];
            }
            for(int i = 0; i < nums.size(); ++i)nums[i] = aux[i];
            exp *= 10;
        }
    }
};

解法2.2 桶排序

间隔为d，则(mathrm{d} geq frac{n\_max - n\_min}{n-1} = b)。因此设置n-1个桶，桶高为(b)，每个桶存储最小和最大的元素，最后依次比较使用过的相邻两个桶的间隔，取最大值

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)return 0;
        int n_max = *max_element(nums.begin(), nums.end());
        int n_min = *min_element(nums.begin(), nums.end());

        int b = max(1, int((n_max - n_min) / (nums.size() - 1)));
        vector<vector<int>>cnt(int((n_max - n_min) / b) + 1, vector<int>(2));

        for(auto &v : cnt){
            v[0] = INT_MAX;
            v[1] = INT_MIN;
        }
        for(int x : nums){
            int idx = floor((x - n_min) / b);
            cnt[idx][0] = min(cnt[idx][0], x);
            cnt[idx][1] = max(cnt[idx][1], x);
        }
        int res = 0;
        int pre_max = cnt[0][1];
        for(int i = 1; i < cnt.size(); ++i){
            if(cnt[i][0] == INT_MAX)continue;
            res = max(res, cnt[i][0] - pre_max);
            pre_max = cnt[i][1];
        }
        return res;
    }
};