C ++程序，用于查找对单词进行分区的方式，以便每个单词都是回文
在这里，我们将讨论一个C++程序，以找到对单词进行分区的多种方法，以使每个单词都是回文。
演算法
Begin
Take the word as input.
Function partitionadd(vector<vector<string> > &u, string &s, vector<string> &tmp, int index):
if (index == 0)
tmp.clear()
for i = index to length-1
st = st + s[i]
if (checkPalin(st))
tmp.push_back(st)
if (i+1 < length)
partitionadd(u, s, tmp, i+1)
else
u.push_back(tmp)
tmp = curr
return
End
Begin
Function partition(string st, vector<vector<string> >&u):
vector<string> tmp
addStrings(u, st, tmp, 0)
printSol(u) //to print the solution
return
End示例
#include <bits/stdc++.h>
using namespace std;
bool checkPalin(string s) { //check if string is palindrome or not.
int length = s.length();
length--;
for (int i=0; i<length; i++) {
if (s[i] != s[length])
return false;
length--;
}
return true;
}
void printSol(vector<vector<string> > part) { //print the solution
for (int i = 0; i < part.size(); ++i) {
for(int j = 0; j < part[i].size(); ++j)
cout << part[i][j] << " ";
cout << endl;
}
return;
}
void partitionadd(vector<vector<string> > &u, string &s, vector<string> &tmp, int index) {
int length = s.length(); //store length of the string
string st;
vector<string> curr = tmp;
//如果当前字符串是回文，则遍历所有索引并递归添加其余分区。
if (index == 0)
tmp.clear();
for (int i = index; i < length; ++i) {
st = st + s[i];
if (checkPalin(st)) {
tmp.push_back(st);
if (i+1 < length)
partitionadd(u,s,tmp,i+1);
else
u.push_back(tmp);
tmp = curr;
}
}
return;
}
void partition(string st, vector<vector<string> >&u) //generate all palindromic partitions of 'str' and stores the result in 'm'. {
vector<string> tmp;
addStrings(u, st, tmp, 0);
printSol(u);
return;
}
int main() {
string s = "tutorials";
vector<vector<string> > part;
cout<<"t分区数："<<endl;
partition(s, part);
return 0;
}输出结果
t分区数： t u t o r i a l s tut o r i a l s