数组中的最小差异 [英] Minimum difference in an array

问题描述

我想要找到一个数组的所有元素之间的最小差异。我阅读了各种其他问题，但无法找到错误的确切来源在以下代码。

I want to find the minimum difference among all elements of an array. I read through various other questions, but couldn't find the exact source of the error in the following code.

#include<iostream>
#include<stdio.h>
#include<cstdlib>

using namespace std;

void quicksort(long int *lp, long int *rp);

int main()
{
    int t,n;
    long int s[5000];

    cin>>t;
    while(t--){
    cin>>n;
    for(int i=0;i<n;i++) cin>>s[i];
    quicksort(&s[0],&s[n-1]);
    //cout<<"passes:"<<passes<<endl;
    //for(int i=0;i<n;i++) cout<<s[i]<<" ";
    long int min = abs(s[1]-s[0]);
    //cout<<endl<<min;
    for(int i=1;i<(n-1);i++){
        long int temp = abs(s[i]-s[i+1]);
        if (temp <= min) min = temp;
    }
    cout<<min;
    }
}

void quicksort(long int *lp,long int *rp){
    int arraysize= (rp-lp)+1;
    if(arraysize > 1){
       long int *pivot = (lp+(arraysize/2));
       long int swap=0;
      long int *orgl = lp;
      long int *orgr = rp;
      while(lp!=rp){
        while (*lp < *pivot) lp++;
        while (*rp > *pivot) rp--;
        if (lp == pivot) pivot=rp;
        else if (rp == pivot) pivot=lp;
        swap = *lp;
        *lp = *rp;
        *rp = swap;
        if((*lp == *pivot) || ( *rp == *pivot)) break;
    }
    quicksort(orgl,pivot-1);
    quicksort(pivot+1,orgr);
}

}

问题陈述在此链接中提供： http://www.codechef.com/problems/HORSES
你能确定我的程序中的错误吗？

The problem statement is given in this link : http://www.codechef.com/problems/HORSES Can you please identify the error in my program ?

推荐答案

快速排序这不是真正保证O（n * logn）你最好使用排序从< algorithm> 。
这个逻辑看起来不错：

You are using C++ so instead of using your custom quicksort which is not really guarantee O(n*logn) you better use sort from <algorithm>. This logic looks good:

    long int min = abs(s[1]-s[0]);
    //cout<<endl<<min;
    for(int i=1;i<(n-1);i++){
        long int temp = abs(s[i]-s[i+1]);
        if (temp <= min) min = temp;
    }

By the way: 
cout<<min; 
Add cout<<min << endl;

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