Algorithms—131.Palindrome Partitioning

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思路：动态规划。。。。

public class Solution {Map<String,List<List<String>>> map=new HashMap<String, List<List<String>>>();    public List<List<String>> partition(String s) {        if (map.get(s)!=null) {        return map.get(s);}        List<List<String>> ans=new ArrayList<List<String>>();        if(isPartition(s)){        List<String> list=new ArrayList<String>();        list.add(s);        ans.add(list);        }        for (int i = 0; i < s.length()-1; i++) {String k=s.substring(0,i+1);if (isPartition(k)) {String t=s.substring(i+1,s.length());List<List<String>> list=partition(t);for (int j = 0; j < list.size(); j++) {List<String> l=new ArrayList<String>();l.add(k);l.addAll(list.get(j));ans.add(l);} }}    map.put(s, ans);    return ans;    }    public boolean isPartition(String s){    if(s.length()==1){    return true;    }    char[] c=s.toCharArray();    int length=c.length;    for (int i = 0; i < length/2; i++) {if(c[i]!=c[length-1-i]){return false;}}    return true;    }}

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