[LeetCode] Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

dp[i][j]=1表示str(i~j)可以是回文。

if dp[i+1][j-1] == 1 && s[i]==s[j]; dp[i][j] = 1;

N^2扫一遍。

然后根据dp[i][j] = 1等价于 (i,j)是一条边。(i,j) 与 (j+1, x)相连。

则就变成了有向图。

然后遍历图，从0->len-1。DFS一遍。

class Solution {
public:
    vector<vector<int> > v;
    int n;
    string org;
    vector<vector<string>> partition(string s) {
        int len = s.size();
        n = len;
        org = s;
        int dp[len][len] ;
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < len; i++) {
            dp[i][i] = 1;
            if (i < len - 1 && s[i] == s[i+1]) dp[i][i+1] = 1;
        }
        
        for (int l = 2; l < len; l++) {
            for (int i = 0; i < len; i++) {
                int j = i+l;
                if (j >= len) continue;
                if (dp[i+1][j-1] == 1 && s[i] == s[j]) dp[i][j] = 1;
            }
        }
        v.clear();
        v.resize(len);
        for (int i = 0; i < len; i++) {
            for (int j = i; j < len; j++) {
                if (dp[i][j] == 1) v[i].push_back(j);
            }
        }
        
        vector<vector<string> > res;
        vector<string> str;
        f(res, str, -1);
        return res;
    }
    
    void f(vector<vector<string> >& res, vector<string>& str, int cur) {
        if (cur == n - 1) {
            res.push_back(str);
            return;
        }
        
        int next = cur+1;
        for (int i = 0; i < v[next].size(); i++) {
            str.push_back(org.substr(next, v[next][i]-next+1));
            f(res, str, v[next][i]);
            str.pop_back();
        }
    }
};

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