LeetCode: Palindrome Partition

LeetCode: Palindrome Partition

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"]
  ]

地址：https://oj.leetcode.com/problems/palindrome-partitioning/
算法：可以用动态规划来解决。用一个二维数组dp来存储所有子问题的解，一维数组dp[i]用来存储0～i的字串的所有解，其中每个解用最后一个回文开始位置来标记。比如对于上面的字符串“aab”，
dp[0]={0},dp[1]={0,1},dp[2]={2}。这样，在构造解的过程中就可以采用递归的方法，对dp[n-1]中的所有值dp[n-1][j]先递归构造出字串0～dp[n-1][j]-1，然后在加上dp[n-1][j]~n-1
字串。具体代码：

 class Solution {
 public:
     vector<vector<string> > partition(string s) {
         int n = s.size();
         if (n < )  return vector<vector<string> >();
         vector<vector<int> > dp(n);
         dp[].push_back();
         for (int i = ; i < n; ++i){
             if(isPalindrome(s.substr(,i+))){
                 dp[i].push_back();
             }
             dp[i].push_back(i);
             for (int j = i-; j >= ; --j){
                 if(isPalindrome(s.substr(j+,i-j))){
                     dp[i].push_back(j+);
                 }
             }
         }
         return constructResult(s,dp,n);
     }
     bool isPalindrome(const string &s){
         int len = s.size();
         int n = len / ;
         int i = ;
         while(i < n && s[i] == s[len--i])    ++i;
         return i == n;
     }
     vector<vector<string> > constructResult(string &s, vector<vector<int> > &dp,int n){
         if (n < ){
             return vector<vector<string> >();
         }
         vector<int>::iterator it = dp[n-].begin();
         vector<vector<string> > result;
         for (; it != dp[n-].end(); ++it){
             if (*it == ){
                 vector<string> temp1;
                 temp1.push_back(s.substr(,n));
                 result.push_back(temp1);
                 continue;
             }
             vector<vector<string> >temp2 = constructResult(s,dp,*it);
             vector<vector<string> >::iterator str_it = temp2.begin();
             for(; str_it != temp2.end(); ++str_it){
                 str_it->push_back(s.substr(*it,n-(*it)));
                 result.push_back(*str_it);
             }
         }
         return result;
     }
 };

第二题：

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.

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

算法：同样用动态规划来解决，但这次只要用一维数组dp来储存所有子问题。其中dp[i]表示字串0～i所用的最小cut，dp[i+1]=min{dp[j-1] | 0 =< j <= i+1 且 字串j~i+1是回文}。代码：

 class Solution {
 public:
     int minCut(string s) {
         int n = s.size();
         if(n < )   return ;
         vector<int> dp(n);
         dp[] = ;
         for(int i = ; i < n; ++i){
             if(isPalindrome(s.substr(,i+))){
                 dp[i] = ;
                 continue;
             }
             int min_value = n;
             for(int j = i-; j >= ; --j){
                 if(dp[j]+ < min_value){
                     if(isPalindrome(s.substr(j+,i-j))){
                         min_value = dp[j] + ;
                     }
                 }
             }
             dp[i] = min_value;
         }
         return dp[n-];
     }
     bool isPalindrome(const string &s){
         int len = s.size();
         int n = len / ;
         int i = ;
         while(i < n && s[i] == s[len--i])    ++i;
         return i == n;
     }
 };