LeetCode :131分割回文串

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。
返回符合要求的 最少分割次数 。
示例 1：
输入：s = “aab”
输出：1
解释：只需一次分割就可将 s 分割成 [“aa”,“b”] 这样两个回文子串。
示例 2：
输入：s = “a”
输出：0
示例 3：
输入：s = “ab”
输出：1
提示：
1 <= s.length <= 2000
s 仅由小写英文字母组成

2023年1月1日解答

class Solution {
public:
int minCut(string s) {
m_c = s.length();
vector<vector> is;
is.assign(m_c, vector(m_c+1));
for (int c = 0; c < m_c; c++)
{
//长度为1的字符串一定是回文
is[c][1] = true;
}
for (int c = 0; c + 1 < m_c; c++)
{
is[c][2] = (s[c] == s[c + 1]);
}

	 for (int len = 3; len <= m_c; len++)
	 {
		 for (int c = 0; c + len - 1 < m_c; c++)
		 {
			 is[c][len] = is[c + 1][len - 2] && (s[c] == s[c + len - 1]);
		 }
	 }

	 //最少多少个回文构成
	 vector<int> dp(m_c + 1,INT_MAX);
	 dp[0] = 0;
	 for (int c = 0; c < m_c; c++)
	 {
		 for (int len = 1; len <= m_c; len++)
		 {
			 if (is[c][len] && (c+len <= m_c ))
			 {
				 dp[c + len] = min(dp[c + len], dp[c] + 1);
			 }
		 }
	 }
	 return dp[m_c] - 1;
 }
 int m_c;

};

2023年8月

//马拉车计算回文回文
class CPalindrome
{
public:
//vOddHalfLen[i]表示 以s[i]为中心，且长度为奇数的最长回文的半长，包括s[i]
//比如：“aba” vOddHalfLen[1]为2 “abba” vEvenHalfLen[1]为2
static void Do(vector& vOddHalfLen, vector& vEvenHalfLen,const string& s)
{
vector v;
for (const auto& ch : s)
{
v.emplace_back(ch);
v.emplace_back(‘*’);
}
v.pop_back();

	const int len = v.size();
	vector<int> vHalfLen(len);
	int center = -1, r = -1;
	//center是对称中心,r是其右边界（闭）
	for (int i = 0; i < len; i++)
	{
		int tmp = 1;
		if (i <= r)
		{
			int pre = center - (i - center);
			tmp = min(vHalfLen[pre], r - i + 1);
		}
		for (tmp++; (i + tmp - 1 < len) && (i - tmp + 1 >= 0) && (v[i + tmp - 1] == v[i - tmp + 1]); tmp++);
		vHalfLen[i] = --tmp;
		const int iNewR = i + tmp - 1;
		if (iNewR > r)
		{
			r = iNewR;
			center = i;
		}
	}

	vOddHalfLen.resize(s.length());
	vEvenHalfLen.resize(s.length());
	for (int i = 0; i < len; i++)
	{
		if (i & 1)
		{
			vEvenHalfLen[i / 2] = vHalfLen[i] / 2;
			
		}
		else
		{
			vOddHalfLen[i / 2] = (vHalfLen[i]+1) / 2 ;				
		}
	}
}

};
class Solution {
public:
int minCut(string s) {
m_c = s.length();
vector vOddHalfLen, vEvenHalfLen;
CPalindrome::Do(vOddHalfLen, vEvenHalfLen,s);
//邻接表
vector<vector> vNeiBo(m_c+1);
for (int i = 0; i < m_c; i++)
{
for (int len = 1; len <= vOddHalfLen[i]; len++)
{
const int cur = i - len + 1;
const int next = i + len;
vNeiBo[cur].emplace_back(next);
}
for (int len = 1; len <= vEvenHalfLen[i]; len++)
{
const int cur = i - len + 1;
const int next = i + len+1;
vNeiBo[cur].emplace_back(next);
}
}

	queue<int> que;
	que.emplace(0);
	vector<int> vDis(m_c+1, -1);
	vDis[0] = 0;
	while (que.size())
	{
		const int cur = que.front();
		que.pop();
		const int curDis = vDis[cur];
		for (const auto& next : vNeiBo[cur])
		{
			if (-1 != vDis[next])
			{
				continue;
			}
			vDis[next] = curDis + 1;
			que.emplace(next);
		}
	}
	return vDis.back() - 1;
}
int m_c;

};