vector, list, map在遍历时删除符合条件的元素实现方法
发布时间:2020-12-16 05:15:36 所属栏目:百科 来源:网络整理
导读:直接看源码,内有详细解释 /*测试vector,list,map遍历时删除符合条件的元素本代码测试环境: ubuntu12 + win7_x64*/#include iostream#include vector#include list#include map#include iteratorusing namespace std;void Remove1(vectorint lt;int::iterato
直接看源码,内有详细解释 /* 测试vector,list,map遍历时删除符合条件的元素 本代码测试环境: ubuntu12 + win7_x64 */ #include <iostream> #include <vector> #include <list> #include <map> #include <iterator> using namespace std; void Remove1(vector<int> &vec,int num) { for (vector<int>::iterator it = vec.begin(); it != vec.end();) { if (*it == num) it = vec.erase(it); else it++; } } void Remove2(list<int> &lst,int num) { list<int>::iterator it; for (it=lst.begin(); it!=lst.end();) { if (*it == num) { lst.erase(it++); } else it++; } } void initMap(map<int,int>& m,int arr[],int arrLen) { for(int i = 0; i < arrLen; i++) m[i] = arr[i]; } void Remove_map(map<int,int num) { map<int,int>::iterator it; for(it = m.begin(); it != m.end();) { if (it->second == num) m.erase(it++); else it++; } } void displayMap(map<int,int>& m) { map<int,int>::iterator it = m.begin(); while(it != m.end()) { cout << "key = " << it->first << ",value = " << it->second << endl; it++; } cout << endl; } int main(void) { int arr[] = {1,3,5,13,7,9}; int arrLen = sizeof(arr) / sizeof(arr[0]); #if 1 // test vector vector<int> vec(arr,arr+arrLen); Remove1(vec,5); copy(vec.begin(),vec.end(),ostream_iterator<int>(cout," ")); cout << endl << endl; #endif // test list list<int> lst(arr,arr+arrLen); Remove2(lst,5); copy(lst.begin(),lst.end()," ")); cout << endl << endl; // test map map<int,int> m; initMap(m,arr,arrLen); Remove_map(m,5); displayMap(m); return 0; } /* Win7_x64运行结果: 1 3 13 7 7 9 1 3 13 7 7 9 key = 0,value = 1 key = 1,value = 3 key = 5,value = 13 key = 6,value = 7 key = 8,value = 7 key = 9,value = 9 Ubuntu12运行结果: [zcm@cpp #54]$make g++ -Wall -Os -DLINUX -o a a.cpp [zcm@cpp #55]$./a 1 3 13 7 7 9 1 3 13 7 7 9 key = 0,value = 9 */ 以上就是小编为大家带来的小编觉得挺不错的,现在就分享给大家,也给大家做个参考。一起跟随小编过来看看吧全部内容了,希望大家多多支持编程小技巧~ (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |