数据结构与算法每日一题——链表(234. 回文链表)

234. 回文链表

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {boolean}
 */
var isPalindrome = function (head) {
    // 解法一:
    // 复制链表值到数组列表中。
    // 使用双指针法判断是否为回文
    // let stack = []
    // while (head) {
    //     stack.push(head.val)
    //     head = head.next
    // }
    // for (let i = 0, j = stack.length - 1; i < j; i++, j--) {
    //     if (stack[i] != stack[j]) {
    //         return false
    //     }
    // }
    // return true
    // 解法二
    // 1:快慢指针同时移动，慢指针移动一步，快指针移动两步；直到快指针为null时，则找到了中位数
    // 2:从中位数的下一位进行反转
    // 3:然后p1指向head p2指向中位数的下一位进行比较，不相等返回false,相等返回true
    const reverseList = (head) => {
        let prev = null;
        let curr = head;
        while (curr !== null) {
            let nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    const endOfFirstHalf = (head) => {
        let fast = head;
        let slow = head;
        while (fast.next !== null && fast.next.next !== null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    var isPalindrome = function (head) {
        if (head == null) return true;

        // 找到前半部分链表的尾节点并反转后半部分链表
        const firstHalfEnd = endOfFirstHalf(head);
        const secondHalfStart = reverseList(firstHalfEnd.next);

        // 判断是否回文
        let p1 = head;
        let p2 = secondHalfStart;
        let result = true;
        while (result && p2 != null) {
            if (p1.val != p2.val) result = false;
            p1 = p1.next;
            p2 = p2.next;
        }

        // 还原链表并返回结果
        firstHalfEnd.next = reverseList(secondHalfStart);
        return result;
    };
};