Given a singly linked list, determine if it is a palindrome.
一开始想用栈,但是试来试去发现写不出来遂放弃,后来想想再不济可以转换成数组然后分别两头扫,但是这样就用了O(n) 的空间,再进一步,可不可以在链表里模拟呢。思前想后发现是可以的,只要把链表的头尾接到一起形成个环,然后头指针每次一动1步,尾指针每次移动 len - 1 步就行了,len 是链表的长度。
这样我实现了算法1:
/*** Definition for singly-linked list.* function ListNode(val) {* this.val = val;* this.next = null;* }*/ /*** @param {ListNode} head* @return {boolean}*/ var isPalindrome = function(head) {if (!head) return true;var tail = head;var len = 0;while (tail && tail.next) {len++;tail = tail.next;}len++;tail.next = head;var ring = len;while (tail.val == head.val) {if (tail === head || ring === 0) return true;head = head.next;ring--;for (var i = 0; i < len - 1; i++) {tail = tail.next;}}return false; };
这个算法是正确的,但是很遗憾TLE 了。头尾相接的做法其实是用取模运算的原理来模拟指针运动,可以运用相同原理适当改写程序而避免头尾相接,算法2:
/*** @param {ListNode} head* @return {boolean}*/ var isPalindrome = function(head) {if (!head) return true;var tail = head;var len = 0;while (tail && tail.next) {len++;tail = tail.next;}len++;var p1 = head;var p2 = head;var k = 1;for (var i = 0; i < len - k; i++) {p2 = p2.next;}while (p1 && p2 && p1.val == p2.val) {if (p1 === p2 || k == len) return true;k++;p2 = head;for (var i = 0; i < len - k; i++) {p2 = p2.next;}p1 = p1.next;}return false; };
算法1,2 的区别就是有卵用和没卵用的区别,但是貌似头尾相接要更简洁一点,但是同样是TLE。
后面看了些提示又想出一个土办法:可以copy 一个链表出来然后反转,然后两个链表一起遍历对比,算法3:
/*** Definition for singly-linked list.* function ListNode(val) {* this.val = val;* this.next = null;* }*/ /*** @param {ListNode} head* @return {boolean}*/ var reverseList = function(head) {if (!head) return null;var prev = null;var now = head;var next = head.nextwhile (now) {now.next = prev;prev = now;now = next;if (next) next = next.next;}return prev; };var copyList = function(head) {var p = new ListNode(-1);var c = p;while (head) {var n = new ListNode(head.val);c.next = n;c = n;head = head.next;}return p.next; }var isPalindrome = function(head) {if (!head) return true;var copy = copyList(head);var rev = reverseList(copy);var p = rev;var q = head;while (p && q) {if (p.val !== q.val) return false;p = p.next;q = q.next;}return true; };
这个是可以accept 的,但是是用了O(n) 额外空间,那么转换成数组或者字符串两头扫八成也是可以通过的。
最后看了一些答案,发现虽然翻转整个链表需要O(n)的额外空间,但是翻转一半的话就不需要,那么只需要把原链表翻转一半,然后两头扫就行,关键是怎么找到一半。
要么先扫一遍得到长度,再扫一遍取一半。但是有更简洁的办法可以一次遍历就找出中点,就是快慢指针。设想一个指针按正常速度遍历链表,另一个指针的速度是他的一半,那么当正常速度指针遍历结束的时候,半速指针刚好在链表的中间,利用这个技巧找到链表中点,然后从中点开始翻转,然后两头扫就可以。算法4:
/*** Definition for singly-linked list.* function ListNode(val) {* this.val = val;* this.next = null;* }*/ /*** @param {ListNode} head* @return {boolean}*/ var reverseList = function(head) {if (!head) return null;var prev = null;var now = head;var next = head.nextwhile (now) {now.next = prev;prev = now;now = next;if (next) next = next.next;}return prev; };var isPalindrome = function(head) {if (!head) return true;var fast = head;var slow = head;var m = 1;while (fast) {if (m === 0) {slow = slow.next;}m = (m + 1) % 2;fast = fast.next;}var tail = reverseList(slow);while (head && tail) {if (head.val !== tail.val) return false;head = head.next;tail = tail.next;}return true;};
利用一个整数变量m 来控制slow 指针的速度,fast 每走两步,slow走一步。然后reverse 另一半,两头扫。整个过程还是很简洁的,Accepted。