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"]
  ]
class Solution {
public:
    bool IsPal(string s)
    {
        int n = s.length();
        int i = 0;
        int j = n-1;
        while(i < j)
        {
            if(s[i++] != s[j--])return false;
        }
        return true;
    }
    
    void findPal(vector<vector<string>> &ret, vector<vector<string>> &v, vector<vector<bool>> &b, int l, int n, vector<string> ss)
    {
        if(l == n)
        {
            ret.push_back(ss);
            return;
        }
        
        string s;
        ss.push_back(s);
        int m = ss.size();
        for(int i = 0; i < n-l; i++)
        {
            if(b[i][l])
            {
                ss[m-1] = v[i][l];
                findPal(ret,v,b,l+i+1,n,ss);
            }
        }
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>>v;
        vector<vector<bool>>pal;
        
        vector<vector<string>>ret;
        int n = s.length();
        if(n==0)return ret;
        for(int i = 1; i <= n; i++)
        {
            vector<string> a;
            vector<bool>b;
            for(int j = 0; j <= n-i; j++)
            {
                string s1;
                s1.assign(s,j,i);
                a.push_back(s1);
                bool t = IsPal(s1);
                b.push_back(t);
                
            }
            v.push_back(a);
            pal.push_back(b);
        }
        vector<string> ss;
        findPal(ret,v,pal,0,n,ss);
        
        return ret;
    }
};


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值