九度OJ 1041 简单排序

题目描述：

You are given an unsorted array of integer numbers. Your task is to sort this array and kill possible duplicated elements occurring in it.

输入：

For each case, the first line of the input contains an integer number N representing the quantity of numbers in this array(1≤N≤1000). Next N lines contain N integer numbers(one number per each line) of the original array.

输出：

For each case ,outtput file should contain at most N numbers sorted in ascending order. Every number in the output file should occur only once.

样例输入：

68 8 7 3 7 7

样例输出：

3 7 8

此题用STL中的set容器再恰当不过，因为set容器内的元素就是不重复的，而且每插入一个数都是按递增的顺序排的，内部是一棵红黑树(平衡检索二叉树)。不过要注意的是最后一个元素的输出最后不要有空格。

#include <iostream>#include <cstring>#include <set>using namespace std;int a[1010];int main(){    int num;    while(cin>>num){        memset(a,0,sizeof(a));        set<int>s;        for(int i=0;i<num;i++){            cin>>a[i];            s.insert(a[i]);        }        set<int>::iterator it;        for(it = s.begin();it!=s.end();it++){            if(it == (--s.end()))                cout<<*it;            else                cout<<*it<<" ";        }        cout<<endl;    }    return 0;}

下面不用set容器来解这道题，思路也很简单，重复元素只记录一个，然后进行插入排序即可。

#include <iostream>#include <cstring>#include <algorithm>using namespace std;int a[1010];int b[1010];int main(){    int num;    while(cin>>num){        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        int k = 0;        for(int i=0;i<num;i++){            cin>>a[i];            bool flag = false;            for(int j=0;j<i;j++){                if(a[i] == a[j]){                    flag = true;                    break;                }            }            if(!flag){                b[k] = a[i];                k++;            }        }        sort(b,b+k);        for(int i=0;i<k-1;i++)            cout<<b[i]<<" ";        cout<<b[k-1]<<endl;    }    return 0;}