LeetCode:Palindrome Partitioning
题目如下：（把一个字符串划分成几个回文子串，枚举所有可能的划分）
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"]
]
分析：首先对字符串的所有子串判断是否是回文，设f[i][j] = true表示以i为起点，长度为j的子串是回文,等于false表示不是回文，那么求f[i][j]的动态规划方程如下：
当j = 1，f[i][j] = true;
当j = 2，f[i][j] = (s[i]==s[i+1]),其中s是输入字符串
当j > 2, f[i][j] = f[i+1][j-2] && (s[i] == s[i+j-1])（即判断s[m..n]是否是回文时：只要s[m+1...n-1]是回文并且s[m] = s[n]，那么它就是回文，否则不是回文）
这一题也可以不用动态规划来求f，可以用普通的判断回文的方式判断每个子串是否为回文。 本文地址
求得f后，根据 f 可以构建一棵树，可以通过DFS来枚举所有的分割方式，代码如下：
class Solution {
public:
vector<vector<string>> partition(string s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector< vector<string> >res;
int len = s.length();
if(len == )return res;
//f[i][j] = true表示以i为起点，长度为j的子串是回文
bool **f = new bool*[len];
for(int i = ; i < len; i++)
{
f[i] = new bool[len+];
for(int j = ; j < len+; j++)
f[i][j] = ;
f[i][] = true;
}
for(int k = ; k <= len; k++)
{
for(int i = ; i <= len-k; i++)
{
if(k == )f[i][] = (s[i] == s[i+]);
else f[i][k] = f[i+][k-] && (s[i] == s[i+k-]);
}
}
vector<string> tmp;
DFSRecur(s, f, , res, tmp);
for(int i = ; i < len; i++)
delete [](f[i]);
delete []f;
return res;
}
void DFSRecur(const string &s, bool **f, int i,
vector< vector<string> > &res, vector<string> &tmp)
{//i为遍历的起点
int len = s.length();
if(i >= len){res.push_back(tmp); return;}
for(int k = ; k <= len - i; k++)
if(f[i][k] == true)
{
tmp.push_back(s.substr(i, k));
DFSRecur(s, f, i+k, res, tmp);
tmp.pop_back();
}
}
};
LeetCdoe:Palindrome Partitioning II
题目如下：（在上一题的基础上，找出最小划分次数）
Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut. 本文地址
算法1：在上一题的基础上，我们很容易想到的是在DFS时，求得树的最小深度即可（遍历时可以根据当前求得的深度进行剪枝），但是可能是递归层数太多，大数据时运行超时，也贴上代码：
class Solution {
public:
int minCut(string s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int len = s.length();
if(len <= )return ;
//f[i][j] = true表示以i为起点，长度为j的子串是回文
bool **f = new bool*[len];
for(int i = ; i < len; i++)
{
f[i] = new bool[len+];
for(int j = ; j < len+; j++)
f[i][j] = ;
f[i][] = true;
}
for(int k = ; k <= len; k++)
{
for(int i = ; i <= len-k; i++)
{
if(k == )f[i][] = (s[i] == s[i+]);
else f[i][k] = f[i+][k-] && (s[i] == s[i+k-]);
}
}
int res = len, depth = ;
DFSRecur(s, f, , res, depth);
for(int i = ; i < len; i++)
delete [](f[i]);
delete []f;
return res - ;
}
void DFSRecur(const string &s, bool **f, int i,
int &res, int &currdepth)
{
int len = s.length();
if(i >= len){res = res<=currdepth? res:currdepth; return;}
for(int k = ; k <= len - i; k++)
if(f[i][k] == true)
{
currdepth++;
if(currdepth < res)
DFSRecur(s, f, i+k, res, currdepth);
currdepth--;
}
}
};
算法2：设f[i][j]是i为起点，长度为j的子串的最小分割次数，f[i][j] = 0时，该子串是回文，f的动态规划方程是：
f[i][j] = min{f[i][k] + f[i+k][j-k] +1} ,其中 1<= k <j
这里f充当了两个角色，一是记录子串是否是回文，二是记录子串的最小分割次数，可以结合上一题的动态规划方程，算法复杂度是O(n^3), 还是大数据超时，代码如下：
class Solution {
public:
int minCut(string s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int len = s.length();
if(len <= )return ;
//f[i][j] = true表示以i为起点，长度为j的子串的最小切割次数
int **f = new int*[len];
for(int i = ; i < len; i++)
{
f[i] = new int[len+];
for(int j = ; j < len+; j++)
f[i][j] = len;
f[i][] = ;
}
for(int k = ; k <= len; k++)
{
for(int i = ; i <= len-k; i++)
{
if(k == && s[i] == s[i+])f[i][] = ;
else if(f[i+][k-] == &&s[i] == s[i+k-])f[i][k] = ;
else
{
for(int m = ; m < k; m++)
if(f[i][k] > f[i][m] + f[i+m][k-m] + )
f[i][k] = f[i][m] + f[i+m][k-m] + ;
}
}
}
int res = f[][len], depth = ;
for(int i = ; i < len; i++)
delete [](f[i]);
delete []f;
return res;
}
};
算法3：同上一题，用f来记录子串是否是回文，另外优化最小分割次数的动态规划方程如下,mins[i] 表示子串s[0...i]的最小分割次数：
- 如果s[0...i]是回文，mins[i] = 0
- 如果s[0...i]不是回文，mins[i] = min{mins[k] +1 (s[k+1...i]是回文) 或 mins[k] + i-k (s[k+1...i]不是回文)} ，其中0<= k < i
代码如下，大数据顺利通过，结果Accept： 本文地址
class Solution {
public:
int minCut(string s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int len = s.length();
if(len <= )return ;
//f[i][j] = true表示以i为起点，长度为j的子串是回文
bool **f = new bool*[len];
for(int i = ; i < len; i++)
{
f[i] = new bool[len+];
for(int j = ; j < len+; j++)
f[i][j] = false;
f[i][] = true;
}
int mins[len];//mins[i]表示s[0...i]的最小分割次数
mins[] = ;
for(int k = ; k <= len; k++)
{
for(int i = ; i <= len-k; i++)
{
if(k == )f[i][] = (s[i] == s[i+]);
else f[i][k] = f[i+][k-] && (s[i] == s[i+k-]);
}
if(f[][k] == true){mins[k-] = ; continue;}
mins[k-] = len - ;
for(int i = ; i < k-; i++)
{
int tmp;
if(f[i+][k-i-] == true)tmp = mins[i]+;
else tmp = mins[i]+k-i-;
if(mins[k-] > tmp)mins[k-] = tmp;
}
}
for(int i = ; i < len; i++)
delete [](f[i]);
delete []f;
return mins[len-];
}
};
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