LeetCode 132. 分割回文串 II（DP）

1. 题目

给定一个字符串 s，将 s 分割成一些子串，使每个子串都是回文串。

返回符合要求的最少分割次数。

示例:
输入: "aab"
输出: 1
解释: 进行一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/palindrome-partitioning-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• dp[i]表示到 i 为止的子串最少需要分割多少次
• 如果一个子串为回文串，dp[i] = 0
• 如果不是，遍历所有的 j （j <= i）,如果s[j,i]是回文串，dp[i] = min(dp[i], dp[j-1]+1)

28 / 29 个通过测试用例

# 超时例子
"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"

class Solution {
public:
    int minCut(string s) {
    	int i,j,n = s.size();
    	vector<int> dp(n,0);
    	if(s.size()<=1)
    		return 0;
    	for(i = 0; i < n; ++i)
            dp[i] = i;
    	for(i = 1; i < n; ++i)
    	{
    		for(j = i; j > 0; --j)
    		{
                if(ispalindrome(s,0,i))
                    dp[i] = 0;
    			else if(ispalindrome(s, j, i))
    				dp[i] = min(dp[i], dp[j-1]+1);
    		}
    	}
    	return dp[n-1];
    }
    bool ispalindrome(string& s, int l, int  r)
    {
    	while(l < r)
    	{
    		if(s[l++]!=s[r--])
    			return false;
    	}
    	return true;
    }
};

• 预先预处理得到所有可能的区间是否是是回文串
• 参考：LeetCode 5. 最长回文子串（动态规划）

class Solution {
public:
    int minCut(string s) {
    	int i,j,len,n = s.size();
    	vector<int> dp(n,0);
        vector<vector<bool>> ispalind(n,vector<bool>(n,false));
    	if(s.size()<=1)
    		return 0;
    	for(i = 0; i < n; ++i)
        {
        	dp[i] = i;
        	ispalind[i][i] = true;
        	if(i < n-1 && s[i]==s[i+1])
                ispalind[i][i+1] = true;
        }
        for(len = 1; len < n; ++len)
        {
            for(i = 0; i < n-len; ++i)
            {
                if(ispalind[i][i+len-1] && i-1>=0 && s[i-1]==s[i+len])//是回文串
                    ispalind[i-1][i+len] = true;
            }
        }
    	for(i = 1; i < n; ++i)
    	{
    		for(j = i; j > 0; --j)
    		{
                if(ispalind[0][i])
                    dp[i] = 0;
    			else if(ispalind[j][i])
    				dp[i] = min(dp[i], dp[j-1]+1);
    		}
    	}
    	return dp[n-1];
    }
};

124 ms 7.4 MB

---

我的CSDN博客地址 https://michael.blog.csdn.net/

长按或扫码关注我的公众号（Michael阿明），一起加油、一起学习进步！