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C++实现LeetCode(24.成对交换节点)

程序员文章站 2022-07-07 23:34:38
[leetcode] 24. swap nodes in pairs 成对交换节点given a linked list, swap every two adjacent nodes and...

[leetcode] 24. swap nodes in pairs 成对交换节点

given a linked list, swap every two adjacent nodes and return its head.

you may not modify the values in the list's nodes, only nodes itself may be changed.

example:

given

1->2->3->4

, you should return the list as

2->1->4->3.

这道题不算难,是基本的链表操作题,我们可以分别用递归和迭代来实现。对于迭代实现,还是需要建立 dummy 节点,注意在连接节点的时候,最好画个图,以免把自己搞晕了,参见代码如下:

解法一:

class solution {
public:
    listnode* swappairs(listnode* head) {
        listnode *dummy = new listnode(-1), *pre = dummy;
        dummy->next = head;
        while (pre->next && pre->next->next) {
            listnode *t = pre->next->next;
            pre->next->next = t->next;
            t->next = pre->next;
            pre->next = t;
            pre = t->next;
        }
        return dummy->next;
    }
};

递归的写法就更简洁了,实际上利用了回溯的思想,递归遍历到链表末尾,然后先交换末尾两个,然后依次往前交换:

解法二:

class solution {
public:
    listnode* swappairs(listnode* head) {
        if (!head || !head->next) return head;
        listnode *t = head->next;
        head->next = swappairs(head->next->next);
        t->next = head;
        return t;
    }
};

解法三:

class solution {
    public listnode swappairs(listnode head) {
        if (head == null || head.next == null) {
            return head;
        }
        listnode newhead = head.next;
        head.next = swappairs(newhead.next);
        newhead.next = head;
        return newhead;
    }
}

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