回文结构:
把中间作为起点,往左和往右的内容相同.
提供了三种解法:
1-额外空间复杂度为O(n) : 创建一个stack,将list中的数据遍历存到stack中,然后判断stack.pop.data是否等于list.next.value.
2-额外空间复杂度为O(n/2): 设置两个指针一个快一个慢,快指针一次走两步,慢指针一次走一步.当快指针不能往下走的时候,满指针到list中间.
将中间后边的部分存到栈里,和方法一 一样去判断.
3-额外空间复杂度为O(1),类似于方法2的办法找到list中间,找到之后,后半部分反转,变为类似于这样的结构1>2>3<4<5.然后两个指针一个指向head,一个指向end,遍历比较.
import java.util.Stack;
public class isPalindrome {
public static void main(String[] args) {
Node n1 = new Node(1);
Node n2 = new Node(2);
n1.next =n2;
Node n3 = new Node(3);
n2.next = n3;
Node n4 = new Node(2);
n3.next = n4;
Node n5 = new Node(1);
n4.next = n5;
System.out.println(isPalindrome3(n1));
}
//need n extra space
public static boolean isPalindrome1(Node head) {
Stack<Node> stack = new Stack<Node>();
Node cur = head;
while(cur != null) {
stack.add(cur);
cur = cur.next;
}
cur = head;
while(!stack.isEmpty()) {
if(stack.pop().data != cur.data) {
return false;
}else {
cur = cur.next;
}
}
return true;
}
//need n/2 extra space
public static boolean isPalindrome2(Node head) {
Node f = head.next;
Node s = head;
while(f.next.next != null && s.next != null) {
f = f.next.next;
s = s.next;
}
Stack<Node> stack = new Stack<Node>();
while(s != null) {
stack.add(s.next);
s = s.next;
}
f = head;
while(!stack.isEmpty()) {
if(stack.pop().data != f.data) {
return false;
}else {
f = f.next;
}
}
return true;
}
//need O(1) extra space
//按照方法2的办法找到list的中间节点,将后半部分的list反转.反转后一个指针从开头,另一个从中间逐个比较.
public static boolean isPalindrome3(Node head) {
Node n1 = head;
Node n2 = head;
while(n2.next != null && n2.next.next != null) {
n1 = n1.next;
n2 = n2.next.next;
}//n1 is mid
n2 = n1.next;
n1.next = null;
Node n3 = null;
//right part convert
while(n2 != null) {
n3 = n2.next; //save next node
n2.next = n1; //contact n2 to mid
n1 = n2;
n2 = n3;
}
//n3 > end
n3 = n1;
n2 = head;
while(n1 != null && n2 != null) {
if(n1.data != n2.data) {
return false;
}else {
n1 = n1.next;
n2 = n2.next;
}
}
return true;
}
}