LeetCode:Palindrome Partitioning,Palindrome Partitioning II
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：
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当j = 1，f[i][j] = true;
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当j = 2，f[i][j] = (s[i]==s[i+1]),其中s是输入字符串
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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：
1 class Solution { 2 public: 3 vector<vector<string>> partition(string s) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 vector< vector<string> >res; 7 int len = s.length(); 8 if(len == 0)return res; 9 //f[i][j] = true表示以i为起点，长度为j的子串是回文 10 bool **f = new bool*[len]; 11 for(int i = 0 ; i < len; i++) 12 { 13 f[i] = new bool[len+1]; 14 for(int j = 0; j < len+1; j++) 15 f[i][j] = 0; 16 f[i][1] = true; 17 } 18 for(int k = 2; k <= len; k++) 19 { 20 for(int i = 0; i <= len-k; i++) 21 { 22 if(k == 2)f[i][2] = (s[i] == s[i+1]); 23 else f[i][k] = f[i+1][k-2] && (s[i] == s[i+k-1]); 24 } 25 } 26 vector<string> tmp; 27 DFSRecur(s, f, 0, res, tmp); 28 for(int i = 0 ; i < len; i++) 29 delete [](f[i]); 30 delete []f; 31 return res; 32 } 33 34 void DFSRecur(const string &s, bool **f, int i, 35 vector< vector<string> > &res, vector<string> &tmp) 36 {//i为遍历的起点 37 int len = s.length(); 38 if(i >= len){res.push_back(tmp); return;} 39 for(int k = 1; k <= len - i; k++) 40 if(f[i][k] == true) 41 { 42 tmp.push_back(s.substr(i, k)); 43 DFSRecur(s, f, i+k, res, tmp); 44 tmp.pop_back(); 45 } 46 47 } 48 };
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：
1 class Solution { 2 public: 3 int minCut(string s) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 int len = s.length(); 7 if(len <= 1)return 0; 8 //f[i][j] = true表示以i为起点，长度为j的子串是回文 9 bool **f = new bool*[len]; 10 for(int i = 0 ; i < len; i++) 11 { 12 f[i] = new bool[len+1]; 13 for(int j = 0; j < len+1; j++) 14 f[i][j] = 0; 15 f[i][1] = true; 16 } 17 for(int k = 2; k <= len; k++) 18 { 19 for(int i = 0; i <= len-k; i++) 20 { 21 if(k == 2)f[i][2] = (s[i] == s[i+1]); 22 else f[i][k] = f[i+1][k-2] && (s[i] == s[i+k-1]); 23 } 24 } 25 int res = len, depth = 0; 26 DFSRecur(s, f, 0, res, depth); 27 for(int i = 0 ; i < len; i++) 28 delete [](f[i]); 29 delete []f; 30 return res - 1; 31 } 32 void DFSRecur(const string &s, bool **f, int i, 33 int &res, int &currdepth) 34 { 35 int len = s.length(); 36 if(i >= len){res = res<=currdepth? res:currdepth; return;} 37 for(int k = 1; k <= len - i; k++) 38 if(f[i][k] == true) 39 { 40 currdepth++; 41 if(currdepth < res) 42 DFSRecur(s, f, i+k, res, currdepth); 43 currdepth--; 44 } 45 46 } 47 48 };
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：
1 class Solution { 2 public: 3 int minCut(string s) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 int len = s.length(); 7 if(len <= 1)return 0; 8 //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 9 int **f = new int*[len]; 10 for(int i = 0 ; i < len; i++) 11 { 12 f[i] = new int[len+1]; 13 for(int j = 0; j < len+1; j++) 14 f[i][j] = len; 15 f[i][1] = 0; 16 } 17 for(int k = 2; k <= len; k++) 18 { 19 for(int i = 0; i <= len-k; i++) 20 { 21 if(k == 2 && s[i] == s[i+1])f[i][2] = 0; 22 else if(f[i+1][k-2] == 0 &&s[i] == s[i+k-1])f[i][k] = 0; 23 else 24 { 25 for(int m = 1; m < k; m++) 26 if(f[i][k] > f[i][m] + f[i+m][k-m] + 1) 27 f[i][k] = f[i][m] + f[i+m][k-m] + 1; 28 } 29 } 30 } 31 int res = f[0][len], depth = 0; 32 for(int i = 0 ; i < len; i++) 33 delete [](f[i]); 34 delete []f; 35 return res; 36 } 37 };
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
1 class Solution { 2 public: 3 int minCut(string s) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 int len = s.length(); 7 if(len <= 1)return 0; 8 //f[i][j] = true表示以i为起点，长度为j的子串是回文 9 bool **f = new bool*[len]; 10 for(int i = 0 ; i < len; i++) 11 { 12 f[i] = new bool[len+1]; 13 for(int j = 0; j < len+1; j++) 14 f[i][j] = false; 15 f[i][1] = true; 16 } 17 int mins[len];//mins[i]表示s[0...i]的最小分割次数 18 mins[0] = 0; 19 for(int k = 2; k <= len; k++) 20 { 21 for(int i = 0; i <= len-k; i++) 22 { 23 if(k == 2)f[i][2] = (s[i] == s[i+1]); 24 else f[i][k] = f[i+1][k-2] && (s[i] == s[i+k-1]); 25 } 26 if(f[0][k] == true){mins[k-1] = 0; continue;} 27 mins[k-1] = len - 1; 28 for(int i = 0; i < k-1; i++) 29 { 30 int tmp; 31 if(f[i+1][k-i-1] == true)tmp = mins[i]+1; 32 else tmp = mins[i]+k-i-1; 33 if(mins[k-1] > tmp)mins[k-1] = tmp; 34 } 35 } 36 for(int i = 0 ; i < len; i++) 37 delete [](f[i]); 38 delete []f; 39 return mins[len-1]; 40 } 41 42 };
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