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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> LeetCode題解:Swap Nodes in Pairs

LeetCode題解:Swap Nodes in Pairs

編輯:C++入門知識

LeetCode題解:Swap Nodes in Pairs


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.

題意:給定一個鏈表,成對交換鄰接的節點

解決思路:每次移動節點向前移動兩位,然後交換就可以了

代碼:

public class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode newHead = new ListNode(0);
        newHead.next = head;

        for(ListNode curr = newHead; curr.next != null && curr.next.next != null; curr = curr.next.next){
            curr.next = swap(curr.next, curr.next.next);
        }

        return newHead.next;
    }

    private ListNode swap(ListNode prev, ListNode after){
        prev.next = after.next;
        after.next = prev;

        return after;
    }
}

 

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