字符串排列数

输入

abbccc

abc (字符分别来自上面一个string)

输出

0,1,3

0,1,4

0,1,5

0,2,3

0,2,4

0,2,5

要求非递归

#include <iostream>#include <cstdio>#include <cstdlib>#include <string.h>#include <Windows.h>#include <string>#include <list>#include <vector>#include <stack>#include <fstream>using namespace std;string s1,s2;vector<vector<unsigned> > charPos;vector<int> sequence;void solve(int index){if (index>=s2.size()){for (int i=0;i<sequence.size();i++){cout << sequence[i] << '/t';}cout << endl;}else{char c=s2[index];for (int i=0;i<charPos[c].size();i++){sequence[index]=charPos[c][i];solve(index+1);}}}int main(){charPos.resize(128);cin >> s1 >> s2;for (int i=0;i<s1.size();i++){charPos[s1[i]].push_back(i);}sequence.resize(s2.size());//recursive//solve(0);//树的每一层上各结点的孩子分枝数是相等的//abbccc,abc//则第一层分枝数1(1个a),第二层分枝数2(2个b),第三层分枝数3(3个c[即对每个b有3个c可选])//non-recursiveint depth=s2.size();//tree depthstack<int,vector<int> > st;//用vector方便查st[i]的值st.push(0);//0 means the frist branch of the rootwhile(1){if (st.size() < depth){st.push(0);}else//全排列形式的树的深度一致，深度达到时输出{//output for (int i=0;i<depth;i++){//即st[i]int b=*(&st.top()-st.size()+1+i);cout << charPos[s2[i]][b] << '/t';}cout << endl;//while(!st.empty())//not empty,loop until finding the node that has branches not visited yet{//has branches not visited yet(0-based)if (st.top()+1 < charPos[s2[st.size()-1]].size()){++st.top();break;}else//return parentst.pop();}if (st.empty())//empty indicates no more branches to search for {//the 1st level of the tree,so terminatebreak;}}}system("pause");return 0;}