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Legal Studies用 C++ 打印字符串的所有回文分区
本例中，我们给定了一个回文串。我们需要打印出此字符串的所有分区。本例中，我们将通过切割字符串来查找字符串的所有可能回文分区。
我们举个例子来理解一下本例 -
输入 - 字符串 = ‘ababa’
输出 - ababa , a bab a, a b a b a ….
对于本例，解决方案是检查子串是否为回文。如果子串为子串，则打印该子串。
示例
以下程序将演示这一解决方案 -
#include<bits/stdc++.h>
using namespace std;
bool isPalindrome(string str, int low, int high){
while (low < high) {
if (str[low] != str[high])
return false;
low++;
high--;
}
return true;
}
void palindromePartition(vector<vector<string> >&allPart, vector<string> &currPart, int start, int n, string str){
if (start >= n) {
allPart.push_back(currPart);
return;
}
for (int i=start; i<n; i++){
if (isPalindrome(str, start, i)) {
currPart.push_back(str.substr(start, i-start+1));
palindromePartition(allPart, currPart, i+1, n, str);
currPart.pop_back();
}
}
}
void generatePalindromePartitions(string str){
int n = str.length();
vector<vector<string> > partitions;
vector<string> currPart;
palindromePartition(partitions, currPart, 0, n, str);
for (int i=0; i< partitions.size(); i++ ) {
for (int j=0; j<partitions[i].size(); j++)
cout<<partitions[i][j]<<" ";
cout<<endl;
}
}
int main() {
string str = "abaaba";
cout<<"Palindromic partitions are :\n";
generatePalindromePartitions(str);
return 0;
}输出
Palindromic partitions are : a b a a b a a b a aba a b aa b a a baab a aba a b a aba aba abaaba
更新于： 14-Jul-2020
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