Palindrome Linked List || Valid Palindrome || Palindrome Partitioning

Palindrome Linked List

Question

Given a singly linked list, determine if it is a palindrome.

Analysis

判断一个链表是否为回文数

• 判断链表节点个数是奇数还是偶数
• 偶数的情况下，两个指针，p1一次走一步，并将路过的元素都压栈，p2一次走两步，当p2为null时，p1刚好指向链表中点，此时p1接着向下，元素与依次弹栈的元素进行对比，假如不等，则返回false，否则返回true
• 奇数的情况下，先将p2先后挪一位，其余相同

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode onestep=head;
        ListNode twostep=head;
        ListNode tmp=head;
        int size=0;
        while(tmp!=null){
            size++;
            tmp=tmp.next;
        }
        boolean result=Findmidnode(onestep,twostep,size);
        return result;
    }
    
    boolean Findmidnode(ListNode tmp1,ListNode tmp2,int size){
        Stack<Integer> stack=new Stack();
        if(size%2==0){
            while(tmp2!=null){
                stack.push(tmp1.val);
                tmp1=tmp1.next;
                tmp2=tmp2.next.next;
            }
        }else{
            tmp2=tmp2.next;
            while(tmp2!=null){
                stack.push(tmp1.val);
                tmp1=tmp1.next;
                tmp2=tmp2.next.next;
            }
            tmp1=tmp1.next;
        }
        while(stack.isEmpty()!=true){
            if(stack.pop()!=tmp1.val)
                return false;
            tmp1=tmp1.next;
        }
        return true;
    }
}

Valid Palindrome

Question

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Analysis

利用StringBuilder将所有符合条件的字符复制至salpha，之后判断salpha大小，利用toUpperCase将所有的字符转换为大写后判断字符是否相等。

Code

public class Solution {
    public boolean isPalindrome(String s) {
        StringBuilder salpha=new StringBuilder();
        for(int i=0;i<s.length();i++){
            if((s.charAt(i)>='a'&&s.charAt(i)<='z')||(s.charAt(i)>='A'&&s.charAt(i)<='Z')||(s.charAt(i)>='0'&&s.charAt(i)<='9'))
                salpha.append(s.charAt(i));
        }
        String str=salpha.toString();
        int size_str=str.length();
        String strupper=str.toUpperCase();
        for(int j=0;j<size_str/2;j++){
            if(strupper.charAt(j)!=strupper.charAt(size_str-j-1))
                return false;
        }
        return true;
    }
}

Palindrome Partitioning

Question

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"]
]

Analysis

DFS

• 从头开始遍历字符串，利用i循环截取不同长度的子串，当截取字符串为回文串时将其加入item，以当前为起点进入下一次调用。
• 当start等于字符串长度的时候，将item加入result，并且调用结束后逐步将item清空。
• List初始化方法：List> res = new ArrayList<>()

Code

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