【刷题-LeetCode】164 Maximum Gap

标签：tps   刷题   存储   lazy   ted   not   cto   and   vector   

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 \(O(n\log n)\)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;
    }
};

【刷题-LeetCode】164 Maximum Gap

标签：tps   刷题   存储   lazy   ted   not   cto   and   vector   

原文地址：https://www.cnblogs.com/vinnson/p/13266649.html