泛型算法之unique
Xredman
posted @ Jun 14, 2009 10:11:32 AM
in 以前博文
, 1138 阅读
#include <iostream>
#include <algorithm>
#include <vector>
#include <list>
using namespace std;
const int N = 10;
int main()
{
int str[8] = {0, 1, 1, 2, 3, 2, 1, 1};
vector<int> vec(str, str + 8);
vector<int>::iterator p, q;
p = vec.begin();
q = unique(vec.begin(), vec.end());
while(p != q)
cout<<*p++<<" ";
cout<<endl;
sort(vec.begin(), vec.end());
p = vec.begin();
q = unique(vec.begin(), vec.end());
while(p != q)
cout<<*p++<<" ";
cout<<endl;
return 0;
}
#include <algorithm>
#include <vector>
#include <list>
using namespace std;
const int N = 10;
int main()
{
int str[8] = {0, 1, 1, 2, 3, 2, 1, 1};
vector<int> vec(str, str + 8);
vector<int>::iterator p, q;
p = vec.begin();
q = unique(vec.begin(), vec.end());
while(p != q)
cout<<*p++<<" ";
cout<<endl;
sort(vec.begin(), vec.end());
p = vec.begin();
q = unique(vec.begin(), vec.end());
while(p != q)
cout<<*p++<<" ";
cout<<endl;
return 0;
}
unique()虽然去掉容器中的重复值，但是只去掉相邻的重复值。
为达到每次只出现单个字母一次的目的，需对其先排序。
unique()操作容器的长度并没有改变，每个独一无二的元素被放到从头开始的下一个自由槽中。
实际上结果为01231223,序列1223是算法的废弃部分，unique()返回一个iterator指向废弃部分的开始处，
典型情况下，这个iterator被传递给相关的容器操作erase()来删除无效元素。
为达到每次只出现单个字母一次的目的，需对其先排序。
unique()操作容器的长度并没有改变，每个独一无二的元素被放到从头开始的下一个自由槽中。
实际上结果为01231223,序列1223是算法的废弃部分，unique()返回一个iterator指向废弃部分的开始处，
典型情况下，这个iterator被传递给相关的容器操作erase()来删除无效元素。
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