palindrome list

dairongpeng · dairongpeng · commit 640ae13f1961 · 2022-02-18T09:51:30.000+08:00
diff --git a/06-链表相关高频题总结.md b/06-链表相关高频题总结.md
@@ -163,221 +163,71 @@ func main() {
 
 > 快慢指针最优解，不使用容器结构(stack)，O(1)：同样的找到中点位置，把右半部分指针回指到中点。接着指针1从L位置，指针2从R位置，往中间遍历。，每步比对，如果有不一样，则不是回文。返回答案之前，把中点右边的指针调整回来
 
-```Java
-package class06;
-
-import java.util.Stack;
-
-public class Code02_IsPalindromeList {
-
-	public static class Node {
-		public int value;
-		public Node next;
-
-		public Node(int data) {
-			this.value = data;
-		}
-	}
-
-	// need n extra space
-	public static boolean isPalindrome1(Node head) {
-	  // 依次进栈
-		Stack<Node> stack = new Stack<Node>();
-		Node cur = head;
-		while (cur != null) {
-			stack.push(cur);
-			cur = cur.next;
-		}
-		// 每个元素和栈中比较
-		while (head != null) {
-			if (head.value != stack.pop().value) {
-				return false;
-			}
-			head = head.next;
-		}
-		return true;
-	}
+```Go
+package main
 
-	// need n/2 extra space
-	// 中点右侧进栈
-	public static boolean isPalindrome2(Node head) {
-		if (head == null || head.next == null) {
-			return true;
-		}
-		Node right = head.next;
-		Node cur = head;
-		while (cur.next != null && cur.next.next != null) {
-			right = right.next;
-			cur = cur.next.next;
-		}
-		Stack<Node> stack = new Stack<Node>();
-		while (right != null) {
-			stack.push(right);
-			right = right.next;
-		}
-		while (!stack.isEmpty()) {
-			if (head.value != stack.pop().value) {
-				return false;
-			}
-			head = head.next;
-		}
-		return true;
-	}
+// 判断一个链表是否是回文链表
+// 解法1：遍历链表，把链表放入一个数组中，倒序遍历数组切片，和正序遍历数组切片依次对比，都相等即为回文，实现略
+// 解法2：遍历链表，按照快慢指针使慢指针定位到链表的中点位置，依次把慢指针指向的值写入数组切片中，此时数组中保存着链表前一半的元素。
+//			且slow的位置是链表的中间位置。按倒序遍历数组切片，与按slow位置顺序遍历链表的元素依次对比，如果都相等，则为回文。实现略
+// 解法2比解法1省了一半的空间，解法一空间复杂度O(n)，解法2空间复杂度为O(n/2)
+// 解法3：不使用额外空间，空间复杂度为O(1)
 
-	// need O(1) extra space
-	// 不使用容器（stack）的方法
-	public static boolean isPalindrome3(Node head) {
-		if (head == null || head.next == null) {
-			return true;
-		}
-		// 慢指针
-		Node n1 = head;
-		// 快指针
-		Node n2 = head;
-		while (n2.next != null && n2.next.next != null) { // find mid node
-			n1 = n1.next; // n1 -> mid
-			n2 = n2.next.next; // n2 -> end
-		}
-		// n1 中点
-		
-		
-		n2 = n1.next; // n2 -> right part first node
-		n1.next = null; // mid.next -> null
-		Node n3 = null;
-		// 右半部逆序指向中点
-		while (n2 != null) { // right part convert
-			n3 = n2.next; // n3 -> save next node
-			n2.next = n1; // next of right node convert
-			n1 = n2; // n1 move
-			n2 = n3; // n2 move
-		}
-		// 引入n3记录最后的位置，之后把右半部再逆序回原来的次序
-		n3 = n1; // n3 -> save last node
-		n2 = head;// n2 -> left first node
-		boolean res = true;
-		while (n1 != null && n2 != null) { // check palindrome
-			if (n1.value != n2.value) {
-				res = false;
-				break;
-			}
-			n1 = n1.next; // left to mid
-			n2 = n2.next; // right to mid
-		}
-		n1 = n3.next;
-		n3.next = null;
-		// 把右半部分再逆序回来
-		while (n1 != null) { // recover list
-			n2 = n1.next;
-			n1.next = n3;
-			n3 = n1;
-			n1 = n2;
-		}
-		return res;
-	}
+type Node struct {
+	Val  int
+	Next *Node
+}
 
-	public static void printLinkedList(Node node) {
-		System.out.print("Linked List: ");
-		while (node != null) {
-			System.out.print(node.value + " ");
-			node = node.next;
-		}
-		System.out.println();
+// IsPalindrome 给定链表头节点，判断该链表是不是回文链表。空间复杂度为O(1)
+func (head *Node) IsPalindrome() bool {
+	// 链表为空，或者链表只有一个节点，是回文结构
+	if head == nil || head.Next == nil {
+		return true
 	}
-
-	public static void main(String[] args) {
-
-		Node head = null;
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(2);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(1);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(2);
-		head.next.next = new Node(3);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(2);
-		head.next.next = new Node(1);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(2);
-		head.next.next = new Node(3);
-		head.next.next.next = new Node(1);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(2);
-		head.next.next = new Node(2);
-		head.next.next.next = new Node(1);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
-		head = new Node(1);
-		head.next = new Node(2);
-		head.next.next = new Node(3);
-		head.next.next.next = new Node(2);
-		head.next.next.next.next = new Node(1);
-		printLinkedList(head);
-		System.out.print(isPalindrome1(head) + " | ");
-		System.out.print(isPalindrome2(head) + " | ");
-		System.out.println(isPalindrome3(head) + " | ");
-		printLinkedList(head);
-		System.out.println("=========================");
-
+	// 慢指针
+	slow := head
+	// 快指针
+	fast := head
+	for fast.Next != nil && fast.Next.Next != nil { // 循环结束，slow停在链表中点位置
+		slow = slow.Next
+		fast = fast.Next.Next
 	}
 
+	fast = slow.Next // 快指针回到中点的下一个节点，之后快指针将从每次走两步，变为每次走一步
+	slow.Next = nil // 从中点截断链表，mid.next -> nil
+	var tmp *Node
+	// 对原链表右半部分，进行逆序，逆序后，从原尾节点指向中点
+	for fast != nil {
+		tmp = fast.Next // tmp暂时保存fast的下一个节点
+		fast.Next = slow // 翻转链表指向
+		slow = fast // slow 移动
+		fast = tmp // fast 移动
+	}
+
+	// tmp指针记录最后的位置，之后把右半部再逆序回原来的次序
+	tmp = slow
+	fast = head
+	var res = true
+	for slow != nil && fast != nil { // 原链表的左右部门进行回文对比
+		if slow.Val != fast.Val {
+			res = false
+			break
+		}
+		slow = slow.Next // 从原链表头节点，往原链表中间节点移动
+		fast = fast.Next // 从原链表尾节点，往原链表中间节点移动
+	}
+	slow = tmp.Next
+	tmp.Next = nil
+	// 把原链表右半部分再逆序回来
+	for slow != nil {
+		fast = slow.Next
+		slow.Next = tmp
+		tmp = slow
+		slow = fast
+	}
+	// 返回回文的判断结果 true
+	return res
 }
-
 ```
 
 ### 1.1.3 面试题二：按值划分单链表
