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Legal Studies用 C++ 列印字串的所有迴文分割槽
本例中，我們給定了一個迴文串。我們需要打印出此字串的所有分割槽。本例中，我們將透過切割字串來查詢字串的所有可能迴文分割槽。
我們舉個例子來理解一下本例 -
輸入 - 字串 = ‘ababa’
輸出 - ababa , a bab a, a b a b a ….
對於本例，解決方案是檢查子串是否為迴文。如果子串為子串，則列印該子串。
示例
以下程式將演示這一解決方案 -
#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;
}輸出
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
更新於： 14-Jul-2020
131 次檢視
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