C language reconstruction [131] split palindrome string

Source code for all topics: Git address

topic

给你一个字符串 s，请你将 s 分割成一些子串，使每个子串都是 回文串 。返回 s 所有可能的分割方案。

回文串 是正着读和反着读都一样的字符串。

 

示例 1：

输入：s = "aab"
输出：[["a","a","b"],["aa","b"]]
示例 2：

输入：s = "a"
输出：[["a"]]
 

提示：

1 <= s.length <= 16
s 仅由小写英文字母组成

Program:

• The idea of ​​retrospective pruning is used, but the code expression is sparse. . . . But it was too late. .

class Solution
{
    
    
public:
    vector<vector<string>> partition(string s)
    {
    
    
        vector<vector<string>> res;
        return part(res, "", s, 0);
    }

    //判断回文
    bool judge(string s, int start, int end)
    {
    
    
        while (start < end)
        {
    
    
            if (s[start] == s[end])
            {
    
    
                start++;
                end--;
            }
            else
                return false;
        }
        return true;
    }
    //分支回溯
    vector<vector<string>> part(vector<vector<string>> last, string last_add, string &s, int start)
    {
    
    
        // 先更新last数组
        if (last_add != "")
        {
    
    
            if (last.size() == 0)
            {
    
    
                vector<string> tmp_add;
                tmp_add.push_back(last_add);
                last.push_back(tmp_add);
            }
            else
            {
    
    
                //字符非空，则先加上
                for (int i = 0; i < last.size(); i++)
                {
    
    
                    last[i].push_back(last_add);
                }
            }
        }
		//如果拆到底了，就返回更新的last
        if (start == s.size())
        {
    
    
            return last;
        }
        //如果没，则开启下一轮
        vector<vector<string>> res;
        for (int i = start; i < s.size(); i++)
        {
    
    
            //判断是否是回文
            if (judge(s, start, i))
            {
    
    
                //如果是则进入下一级
                vector<vector<string>> tmp = part(last, s.substr(start, i - start + 1), s, i + 1);
                
                //然后加入列表
                res.insert(res.end(), tmp.begin(), tmp.end());
            }
            //不是则加长
        }
        return res;
    }
};

Complexity calculation

• Time complexity: O(n2)
• Space complexity: O(n2)