LeetCode Palindrome Partitioning

· 194 words · 1 minute read
2017年7月
LeetCode

131.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"]
]

方法

个人认为所有回文串的方法都可以用dfs或者dp的方式解决

本题在以dfs的框架下加上判断是否为回文串，若是回文串则执行相应的操作。

public class Solution {
    List<List<String>> list = new ArrayList<List<String>>();
    public List<List<String>> partition(String s) {
        if(s == null || s.length() == 0){
            return new ArrayList<List<String>>();
        }
        dfs(s,0,new ArrayList<String>());
        return list;
    }
    public void dfs(String s, int start, ArrayList<String> arrayList){
        if(start >= s.length()){
            list.add(new ArrayList<String>(arrayList));
            return;
        }
        for(int i = start; i < s.length(); i++){
            if(isPal(s,start,i)){
                if(start == i){
                    arrayList.add(Character.toString(s.charAt(start)));
                }else{
                    arrayList.add(s.substring(start,i+1));
                }
                dfs(s, i+1, arrayList);
                arrayList.remove(arrayList.size()-1);
            }
        }
    }
    public boolean isPal(String s, int start, int end){
        while(start < end){
            if(s.charAt(start) != s.charAt(end)){
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
}
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