【LeetCode笔记】Palindrome Partitioning

2 DP methods are used in this implementation:

1. DP method to find all palindromes in string s:

This one is talk thoroughly in before...

2. DP method to find all partitions in string s:

If sub-string s[i..j] is a palindrome, then all partitions for s[0..j] is all partitions for s[0..i-1] + palindromes[i..j]

Code with explanations:

vector<vector<string> > partition(string s){int n = s.size();//pal[i][j]: substring s[i..j] is a palindrome or notvector<vector<bool> > pal(n, vector<bool>(n, false)); //solution[i + 1]: all partitions for substring s[0..i]vector<vector<vector<string> > > solution(n + 1); //solution[0] is a empty matrix: partition for substrings[0..-1]solution[0].push_back(vector<string>(0)); for(int j = 0; j < n; ++j)for(int i = 0; i <= j; ++i)if(s[i] == s[j] && (j - i < 2 || pal[i + 1][j - 1])){pal[i][j] = true;//DP method: all partitions for s[0..j]: all partitions for s[0..i-1] + palindromes[i..j]for(int k = 0; k < solution[i].size(); ++k){ //solution[i]: all partitions for s[0..i-1]solution[i][k].push_back(s.substr(i, j - i + 1)); //add palindromes[i..j]solution[j + 1].push_back(solution[i][k]); //solution[j + 1]: all partitions for s[0..j]solution[i][k].pop_back(); //be sure to recover solution[i] for later use}}return solution[n];}