linear-list/linked-list/swap-nodes-in-pairs

Swap Nodes in Pairs

描述

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

For example, Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

分析

无

代码

// Swap Nodes in Pairs
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        for(ListNode prev = dummy, cur = prev.next, next = cur.next;
                next != null;
                prev = cur, cur = cur.next, next = cur != null ? cur.next: null) {
            prev.next = next;
            cur.next = next.next;
            next.next = cur;
        }
        return dummy.next;
    }
}

// Swap Nodes in Pairs
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        if (head == nullptr || head->next == nullptr) return head;
        ListNode dummy(-1);
        dummy.next = head;

        for(ListNode *prev = &dummy, *cur = prev->next, *next = cur->next;
                next;
                prev = cur, cur = cur->next, next = cur ? cur->next: nullptr) {
            prev->next = next;
            cur->next = next->next;
            next->next = cur;
        }
        return dummy.next;
    }
};

下面这种写法更简洁，但题目规定了不准这样做。

// Swap Nodes in Pairs
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode p = head;

        while (p != null && p.next != null) {
            int tmp = p.val;
            p.val = p.next.val;
            p.next.val = tmp;

            p = p.next.next;
        }

        return head;
    }
}

// Swap Nodes in Pairs
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* p = head;

        while (p && p->next) {
            swap(p->val, p->next->val);
            p = p->next->next;
        }

        return head;
    }
};

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