🌕 剑指 Offer 06. 从尾到头打印链表

难度: 🌕

问题描述

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解法 1 - 递归反转链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    int len = 0;
    public int[] reversePrint(ListNode head) {
        // 思路：
        // 递归反转链表
        ListNode cur = mySol(null, head);
        int[] res = new int[len];
        for(int i = 0; i < len; i ++) {
            res[i] = cur.val;
            cur = cur.next;
        }
        return res;
    }

    private ListNode mySol(ListNode pre, ListNode cur) {
        // 递归终止条件
        if(cur == null) {
            return pre;
        }
        // cur != null
        len ++;
        ListNode next = cur.next;
        cur.next = pre;
        return mySol(cur, next);
    }
}

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输出 1

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解法 2 - 迭代反转链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    int len = 0;
    public int[] reversePrint(ListNode head) {
        // 思路：
        // 先反转链表，同时记录元素个数，再顺序遍历
        ListNode cur = mySol(head);
        int[] res = new int[len];
        int index = 0;
        while(cur != null) {
            res[index] = cur.val;
            index ++;
            cur = cur.next;
        }
        return res;
    }

    private ListNode mySol(ListNode head) {
        if(head == null) {
            return head;
        }
        ListNode pre = null;
        ListNode cur = head;
        while(cur != null) {
            len ++;
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

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输出 2

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解法 3 - 顺序存入 List

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] reversePrint(ListNode head) {
        // 思路：
        // 一次遍历，存入 ArrayList
        ArrayList<Integer> list = new ArrayList<>();
        ListNode cur = head;
        while(cur != null) {
            list.add(cur.val);
            cur = cur.next;
        }
        int len = list.size();
        int[] res = new int[len];
        for(int i = 0; i < len; i ++) {
            res[i] = list.get(len - 1 - i);
        }
        return res;
    }
}

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输出 3