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：
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 };
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：
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 };
