算法笔记 --- 相邻两数最大差值

有一个整形数组A，请设计一个复杂度为O(n)的算法，算出排序后相邻两数的最大差值。

给定一个int数组A和A的大小n，请返回最大的差值。保证数组元素多于1个。

#include <iostream>
#include <vector>
using namespace std;

class Gap {
public:
    int maxGap(vector<int> A, int n) {
        int var_max = A[0], var_min = A[0];
        for(int i = 1; i < n; i++){
            if(var_max < A[i])
                var_max = A[i];
            if(var_min > A[i])
                var_min = A[i];
        }
        if(var_max == var_min)
            return 0;
            
        double var_unit = (double)(var_max - var_min)/(double)n;
        // vector<int> bucket[n+1];
        vector<vector<int> > bucket(n+1);
        for(int i = 0; i < n; i++){
            bucket[getIndex(var_unit, A[i], var_min)].push_back(A[i]);
        }
        int res = -1, tmp;
        int index1 = 0, index2 = 1;
        while(index2 < n+1){
            // cout<<"index1: "<<index1<<"|| index2: "<<index2<<endl;
            if(!bucket[index1].empty() && !bucket[index2].empty()){
                tmp = getMin(bucket[index2]) - getMax(bucket[index1]);
                index1++, index2++;
                if(tmp > res)
                    res = tmp;
            }
                
            if(bucket[index1].empty())
                index1++;
            if(bucket[index2].empty())
                index2++;
        }
        return res;
    }
    int getIndex(double var_unit, int var, int var_min){
        return ((double)(var-var_min)/var_unit);
    }
    int getMax(vector<int> A){
        int tmp = A[0];
        for(int i = 1; i < A.size(); i++){
            if(tmp < A[i])
                tmp = A[i];
        }
        return tmp;
    }
    int getMin(vector<int> A){
        int tmp = A[0];
        for(int i = 1; i < A.size(); i++){
            if(tmp > A[i])
                tmp = A[i];
        }
        return tmp;
    }
};
int main()
{

    vector<int> A;
    //3429,6401,8559,1052,4775,6220,3593,2406,4995
    A.push_back(3429), A.push_back(6401), A.push_back(8559), A.push_back(1052), A.push_back(4775);
    A.push_back(6220), A.push_back(3593), A.push_back(2406), A.push_back(4995);
    Gap sorter;
    int res = sorter.maxGap(A, 9);

    cout<<res<<endl;
    
    return 0;
}