leetcode Palindrome Partitioning-优快云博客

本文详细解析了如何使用动态规划和DFS算法解决Palindrome Partitioning问题，并提供了代码实现。通过实例，深入理解如何将字符串分割成回文子串。

Palindrome Partitioning 原题地址：

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

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"]
  ]

之前有道动态规划的题，加一个dfs就可以。

public class Solution {
    private List<List<String>> plist = new ArrayList<List<String>>();
	public List<List<String>> partition(String s) {
		if (s == null || s.length() == 0)
			return plist;
		int len = s.length();
		boolean[][] palin = new boolean[len][len];
		for (int h = 0; h < len; h++)
			for (int i = 0; i < len-h; i++) {
				if (s.charAt(i) == s.charAt(i+h) && (h < 2 || palin[i+1][i+h-1]))
					palin[i][i+h] = true;
			}
		LinkedList<String> list = new LinkedList<String>();
		dfs(palin, list, s, 0);
		return plist;
	}
	
	private void dfs(boolean[][] palin, LinkedList<String> list, String s, int v) {
		if (v == s.length()) {
			List<String> _list = (List<String>) list.clone();
			plist.add(_list);
			return;
		}
		for (int w = v; w < s.length(); w++) {
			if (palin[v][w]) {
				String temp = s.substring(v, w+1);
				list.add(temp);
				dfs(palin, list, s, w+1);
				list.pollLast();
			}
		}
	}
}