Palindrome Partitioning

最新推荐文章于 2021-03-14 00:46:08 发布

最新推荐文章于 2021-03-14 00:46:08 发布 · 86 阅读

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本文介绍了一种递归算法，用于将输入字符串分割成所有可能的子串组合，确保每个子串都是回文串。通过深度优先搜索(DFS)遍历所有分割可能性，并使用辅助函数检查子串是否为回文。

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"]
  ]

public class Solution {
    public List<List<String>> partition(String s) {
    	List<String> item = new ArrayList<String>();
    	List<List<String>> res = new ArrayList<List<String>>();
		if (s == null || s.length() == 0) {
			return res;
		}
		dfs(s,0,item,res);
		return res;
    }

	private void dfs(String s, int start, List<String> item, List<List<String>> res) {
		// TODO Auto-generated method stub
		if (start == s.length()) {
			res.add(new ArrayList<String>(item));
			return;
		}
		for (int i = start; i < s.length(); i++) {
            String str = s.substring(start, i+1);
            if (isPalindrome(str)) {
                item.add(str);
                dfs(s, i+1, item, res);
                item.remove(item.size() - 1);
            }
        }
	}

	private boolean isPalindrome(String s) {
		int low = 0;
        int high = s.length()-1;
        while(low < high){
            if(s.charAt(low) != s.charAt(high)) {
            	return false;
            }
            low++;
            high--;
        }
        return true;
	}
}