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.
1) 初次尝试：在解决这个问题的时候，我第一反映是用贪婪算法，从头到尾遍历s，找出回文，但是事实上这样得不到最少的分割。例如：
ababbbabbaba
贪心算法得到的结果将是：aba,bbb,abba,b,a
函数返回的结果是4，即需要4次分割。但是更少的分割：
aba,b,bbabb,aba
需要三次分割。
2) 重新思考这个问题，翻阅了“算法导论”的动态规划章节，发现这个求最优解的问题也可以用动态规划的方法解决。
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input：
x[0...n-1],
c(x, i):
minimum cuts needed for a palindrome partitioning of x[0...i], initialize to be MAX.INT
recursion:
c(x, i) = min{c(x, j) + 1, c(x, i)}, 0 ≤ i < j ≤ n-1, if c[i...j] is a palindrome
return:
c(x, n-1): minimum cuts needed for a palindrome partitioning of x, also the return value,
dp[i, j]:
= true, if x[i...j] is palindrome; = false, else
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package Palindrome.Partitioning.II;
public class net_ex1 {
public int minCut(String s) {
// Start typing your Java solution below
// DO NOT write main() function
int n = s.length();
// cut[i]表示s[i..n-1]需要的最小的切割
// p[i][j]表示s[i..j]是否是回文
int cut[] = new int[n + 1];
boolean p[][] = new boolean[n][n];
for (int i = 0; i < n; i++) {
cut[i] = Integer.MAX_VALUE;
p[i][i] = true;// 肯定是回文
}
cut[n] = -1;
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
p[i][j] = ((s.charAt(i) == s.charAt(j)) && (j - i + 1 <= 2 || p[i + 1][j - 1]));// 判断回文，也是一个递归
if (p[i][j]) {
cut[i] = Math.min(cut[i], cut[j + 1] + 1);
}
}
}
return cut[0];
}
public static void main(String args[]) {
String t1 = "adabdcaebdcebdcacaaaadbbcadabcbeabaadcbcaaddebdbddcbdacdbbaedbdaaecabdceddccbdeeddccdaabbabbdedaaabcdadbdabeacbeadbaddcbaacdbabcccbaceedbcccedbeecbccaecadccbdbdccbcbaacccbddcccbaedbacdbcaccdcaadcbaebebcceabbdcdeaabdbabadeaaaaedbdbcebcbddebccacacddebecabccbbdcbecbaeedcdacdcbdbebbacddddaabaedabbaaabaddcdaadcccdeebcabacdadbaacdccbeceddeebbbdbaaaaabaeecccaebdeabddacbedededebdebabdbcbdcbadbeeceecdcdbbdcbdbeeebcdcabdeeacabdeaedebbcaacdadaecbccbededceceabdcabdeabbcdecdedadcaebaababeedcaacdbdacbccdbcece";
net_ex1 n = new net_ex1();
System.out.println(n.minCut(t1));
}
}