leetcode]Palindrome Partitioning IIMar

 

Palindrome Partitioning IIMar 

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));
	}

}