Leetcode: palindrome Partition

Leetcode: palindrome Partition

Given a stringS, PartitionSSuch that every substring of the partition is a palindrome.

Return all possible palindrome partitioningS.

For example, givenS="aab",
Return

[["AA", "B"], ["A", "A", "B"]
Address: https://oj.leetcode.com/problems/palindrome-partitioning/
Algorithm: it can be solved through dynamic planning. A two-dimensional array DP is used to store solutions to all sub-problems. The one-dimensional array DP [I] is used to store 0 ~ All solutions of the I string, each of which is marked by the start position of the last input. For example, for the above string "AAB ",
DP [0] = {0}, DP [1] = {0, 1}, DP [2] = {2 }. In this way, the recursive method can be used to construct the 0 ~ string for all the values of DP [n-1] [n-1] [J] In DP [n-1 ~ DP [n-1] [J]-1, then add DP [n-1] [J] ~ N-1
String. Code:

 1 class Solution { 2 public: 3     vector<vector<string> > partition(string s) { 4         int n = s.size(); 5         if (n < 1)  return vector<vector<string> >(); 6         vector<vector<int> > dp(n); 7         dp[0].push_back(0); 8         for (int i = 1; i < n; ++i){ 9             if(isPalindrome(s.substr(0,i+1))){10                 dp[i].push_back(0);11             }12             dp[i].push_back(i);13             for (int j = i-2; j >= 0; --j){14                 if(isPalindrome(s.substr(j+1,i-j))){15                     dp[i].push_back(j+1);16                 }17             }18         }19         return constructResult(s,dp,n);20     }21     bool isPalindrome(const string &s){22         int len = s.size();23         int n = len / 2;24         int i = 0;25         while(i < n && s[i] == s[len-1-i])    ++i;26         return i == n;27     }28     vector<vector<string> > constructResult(string &s, vector<vector<int> > &dp,int n){29         if (n < 1){30             return vector<vector<string> >();31         }32         vector<int>::iterator it = dp[n-1].begin();33         vector<vector<string> > result;34         for (; it != dp[n-1].end(); ++it){35             if (*it == 0){36                 vector<string> temp1;37                 temp1.push_back(s.substr(0,n));38                 result.push_back(temp1);39                 continue;40             }41             vector<vector<string> >temp2 = constructResult(s,dp,*it);42             vector<vector<string> >::iterator str_it = temp2.begin();43             for(; str_it != temp2.end(); ++str_it){44                 str_it->push_back(s.substr(*it,n-(*it)));45                 result.push_back(*str_it);46             }47         }48         return result;49     }50 };

Question 2:

Given a stringS, PartitionSSuch that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioningS.

For example, givenS="aab",
Return1Since the palindrome partitioning["aa","b"]Cocould be produced using 1 cut.

Address: https://oj.leetcode.com/problems/palindrome-partitioning-ii/

Algorithm: Dynamic Programming is also used to solve the problem, but this time we only need to use one-dimensional array DP to store all sub-problems. DP [I] indicates the string 0 ~ The minimum cut used by I, DP [I + 1] = min {DP [J-1] | 0 = <j <= I + 1 and the string J ~ I + 1 is a reply }. Code:

 1 class Solution { 2 public: 3     int minCut(string s) { 4         int n = s.size(); 5         if(n < 1)   return 0; 6         vector<int> dp(n); 7         dp[0] = 0; 8         for(int i = 1; i < n; ++i){ 9             if(isPalindrome(s.substr(0,i+1))){10                 dp[i] = 0;11                 continue;12             }13             int min_value = n;14             for(int j = i-1; j >= 0; --j){15                 if(dp[j]+1 < min_value){16                     if(isPalindrome(s.substr(j+1,i-j))){17                         min_value = dp[j] + 1;18                     }19                 }20             }21             dp[i] = min_value;22         }23         return dp[n-1];24     }25     bool isPalindrome(const string &s){26         int len = s.size();27         int n = len / 2;28         int i = 0;29         while(i < n && s[i] == s[len-1-i])    ++i;30         return i == n;31     }32 };

 

Leetcode: palindrome Partition