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"]
]

递归求解,没什么难度。

public class Solution {
List list = new ArrayList<ArrayList<String>>();
char[] word ;
String ss;
String[] result; public List<List<String>> partition(String s) {
int len = s.length();
if( len == 0)
return list;
if( len == 1 ){
ArrayList<String> l1 = new ArrayList<String>();
l1.add(s);
list.add(l1);
return list;
}
ss = s;
word = s.toCharArray();
result = new String[len];
for( int i = 0;i < len;i++){
if( isPalindrome(0,i) ){
result[0] = s.substring(0,i+1);
helper(i+1,1);
}
}
return list; } public void helper(int start,int num){ if( start == word.length ){
ArrayList ll = new ArrayList<String>();
for( int i = 0;i<num;i++)
ll.add(result[i]);
list.add(ll);
return ;
} for( int i = start; i < word.length;i++){
if( isPalindrome(start,i) ){
result[num] = ss.substring(start,i+1);
helper(i+1,num+1);
}
} } public boolean isPalindrome(int start,int end){ while( start < end ){
if( word[start] == word[end] ){
start++;
end--;
}else
return false;
}
return true; }
}
 

leetcode 131. Palindrome Partitioning----- java的更多相关文章

  1. [LeetCode] 131. Palindrome Partitioning 回文分割

    Given a string s, partition s such that every substring of the partition is a palindrome. Return all ...

  2. leetcode 131. Palindrome Partitioning 、132. Palindrome Partitioning II

    131. Palindrome Partitioning substr使用的是坐标值,不使用.begin()..end()这种迭代器 使用dfs,类似于subsets的题,每次判断要不要加入这个数 s ...

  3. Java for LeetCode 131 Palindrome Partitioning

    Given a string s, partition s such that every substring of the partition is a palindrome. Return all ...

  4. Leetcode 131. Palindrome Partitioning

    Given a string s, partition s such that every substring of the partition is a palindrome. Return all ...

  5. [leetcode]131. Palindrome Partitioning字符串分割成回文子串

    Given a string s, partition s such that every substring of the partition is a palindrome. Return all ...

  6. Leetcode 22. Generate Parentheses Restore IP Addresses (*) 131. Palindrome Partitioning

    backtracking and invariant during generating the parathese righjt > left  (open bracket and cloas ...

  7. Java for LeetCode 132 Palindrome Partitioning II

    Given a string s, partition s such that every substring of the partition is a palindrome. Return the ...

  8. 【LeetCode】131. Palindrome Partitioning 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 回溯法 日期 题目地址:https://leetco ...

  9. leetcode 132. Palindrome Partitioning II ----- java

    Given a string s, partition s such that every substring of the partition is a palindrome. Return the ...

  10. 【LeetCode】131. Palindrome Partitioning

    Palindrome Partitioning Given a string s, partition s such that every substring of the partition is ...

随机推荐

  1. fragment 添加menu

    http://bbs.51cto.com/thread-1091458-1-1.html 有详解 @Override public void onCreate(Bundle savedInstance ...

  2. Word2013可以写博客

    步骤如下http://www.cnblogs.com/guyichang/p/4629211.html

  3. STM32之RTC配置与初始化-rtc.h rtc.c

    <rtc.h> #include "stm32f10x.h" #ifndef _RTC_H #define _RTC_H typedef struct { vu8 ho ...

  4. Centos6升级内核2.6到3.x过程

    最近公司有一个应用,安装需要内核版本3.1以后,不得已,需要升级下内核版本: 1. 安装必要依赖 # yum groupinstall "Development Tools" #y ...

  5. JS创建自定义对象

    普通对象的创建: 创建对象: 1.people = new Object(); people.name = "lin"; people.age = "26“; 2.创建字 ...

  6. 给伪类设置z-index= -1;

    .column{ position: relative; float: left; padding: 30px 0; width: 25%; z-index: 0; background-color: ...

  7. 深入C#数据类型小部分第二章

    值类型和引用类型C#的值类型包括:结构体(数值类型,bool型,用户定义的结构体),枚举,可空类型. C#的引用类型包括:数组,用户定义的类.接口.委托,object,字符串. 数组的元素,不管是引用 ...

  8. WebGrid Enterprise免费下载

    WebGrid.NET Enterprise是一个为ASP.NET平台下WEB开发而设计的高级数据表格控件.WebGrid.NET为复杂的分层次导航交互式企业级信息传输提供了全面而先进的功能,它允许用 ...

  9. R——启程——豆瓣影评分析

    专业统计的我,自然免不了学R的,今天仔细看了这篇教程(感谢学姐的推荐@喜欢算法的女青年),就学着用R仿照着做一个,作为R语言学习的起点吧. 影评数据是用python爬的,之后会在python爬虫系列补 ...

  10. 夜黑风高的夜晚用SQL语句做了一些想做的事·······

         IT这条漫漫长路注定是孤独的,陪伴我们的只有那些不知冷暖的代码语句和被手指敲打的磨掉了键上的标识的键盘. 之所以可以继续坚持下去,是因为心中有一份永不熄灭的激情. 成功的路上让我们为自己带盐 ...