LeetCode-palindrome-partitioning

最新推荐文章于 2019-05-14 22:43:23 发布

六月二十七

最新推荐文章于 2019-05-14 22:43:23 发布

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分类专栏： LeetCode 文章标签： palindrome-partitioning LeetCode

本文链接：https://blog.youkuaiyun.com/weixin_42146769/article/details/88831535

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本文介绍了一种用于字符串处理的算法，该算法能够将输入的字符串s分割成多个子串，确保每个子串都是回文串，并返回所有可能的回文分割组合。通过递归和回溯的方法实现，适用于字符串分析和模式匹配等问题。

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"]
  ]

import java.util.ArrayList;
import java.util.List;
 
public class Solution {
  public static ArrayList<ArrayList<String>> partition(String s) {
    ArrayList<ArrayList<String>> lists = new ArrayList<>();
    ArrayList<String> list = new ArrayList<>();
    partitionHepler(lists, list, s);
    return lists;
  }
 
  public static void partitionHepler(ArrayList<ArrayList<String>> lists, ArrayList<String> list, String s) {
    if (null == s || s.length() == 0) {
      lists.add(new ArrayList<>(list));
      return;
    }
    int len = s.length();
    for (int i = 0; i <= len;i++) {
      String subStr = s.substring(0, i);
      if (isPalindrome(subStr)) {
        list.add(subStr);
        partitionHepler(lists, list, s.substring(i, len));
        list.remove(list.size() - 1);
      }
    }
  }
 
  public static boolean isPalindrome(String s) {
    if (null == s || s.length() == 0) return false;
    int length = s.length();
    int middle = length / 2;
    for (int i = 0; i < middle;i++) {
      if (s.charAt(i) != s.charAt(length - 1 - i)) {
        return false;
      }
    }
    return true;
  }
}