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131. 分割回文串——回溯+记忆化搜索

爪哇驿站 2022-01-31 阅读 33
class Solution {
public:
    vector<vector<string>> ans; //答案
    vector<vector<bool>> flag;  //i, j表示i - j是否是回文串
    vector<string> temp;    //子串
    vector<vector<string>> partition(string s) {
        int n = s.size();
        flag.assign(n, vector<bool>(n, false));  //赋值,默认为true
        backtrack(s, 0, n);
        return ans;
    }
    bool isPalindrome(const string& s, int i, int j) {  //记忆化搜索
        if(flag[i][j])
            return true;
        if(i >= j)  //"aa"这种情况会出错
            return flag[i][j] = true;
        return flag[i][j] = (s[i] == s[j] && isPalindrome(s, i + 1, j - 1));    //i == j并且 i+1 - j-1 是回文数
    }
    void backtrack(const string& s, int i, int n){
        if(i == n){
            ans.push_back(temp);
            return;
        }
        for(int j = i; j < n; ++j){
            if(isPalindrome(s, i, j)){
                temp.push_back(s.substr(i, j - i + 1));
                backtrack(s, j + 1, n);
                temp.pop_back();
            }
        }
    }
};

Accepted
32/32 cases passed (100 ms)
Your runtime beats 84.45 % of cpp submissions
Your memory usage beats 74.6 % of cpp submissions (73.9 MB)

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