https://oj.leetcode.com/problems/palindrome-partitioning/

给定一个字符串 s,求所有的子串组合,每个子串都是回文的。

比如,aba: {a,b,a},{aba}

对于这类问题(对一个串的划分)基本上就是用递归。

首先设一个 record 数组,记录中间结果,省的多次计算。

class Solution {
public:
vector<vector<string> > partition(string s) {
vector<vector<string> > ans;
if(s.empty())
return ans;
const size_t len = s.size(); vector<vector<bool> > flag;
flag.resize(len);
for(int i = ;i<len;i++)
flag[i].resize(len); for(int i = ;i<len;i++)
for(int j = i;j<len;j++)
{
flag[i][j] = isPal(s,i,j);
} vector<string> ansPiece;
sub(,flag,ans,ansPiece,s);
return ans;
}
void sub(int begin,vector<vector<bool> > &flag,vector<vector<string> > &ans,vector<string> &ansPiece,string &s)
{
if(begin == s.size())
{
vector<string> _ansPiece = ansPiece;
ans.push_back(_ansPiece);
return ;
} //here i means end position
for(int i = begin;i<flag.size();i++)
{
string tempstr;
if(flag[begin][i])
{
tempstr = s.substr(begin,i-begin+);
ansPiece.push_back(tempstr);
sub(i+,flag,ans,ansPiece,s);
ansPiece.pop_back();
}
}
}
bool isPal(string s,int i,int j)
{
if(i==j)
return true; int temp = ;
while(i+temp<j-temp)
{
if(s[i+temp]!=s[j-temp])
return false;
temp++;
}
return true;
}
};

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