May LeetCoding Challenge16 之链表重组

本题两种解法：

1.分别申请odd, even两个头结点。用count来区分奇数偶数。head在链表上后移进行遍历，如果为奇数，将结点添加到odd中，将odd末尾置为null，如果为偶数，将结点添加到 even中，将末尾置为null（为了避免出现环），count+1。最后将even添加到odd末尾，返回odd。此解法没有改变原链表结构，空间复杂度高。

2.在链表上直接进行操作，记录偶数结点开始的位置。然后对链表进行断开连接。

            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;

JAVA

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        ListNode odd = new ListNode(0);
        ListNode even = new ListNode(0);
        ListNode p = odd;
        ListNode q = even;
        int count = 1;
        while(head != null){
            if(count % 2 == 1){
                p.next = head;
                head = head.next;
                p = p.next;
                p.next = null;
            }
            else{
                q.next = head;
                head = head.next;
                q = q.next;
                q.next = null;
            }
            count ++;
        }
        p.next = even.next;
        return odd.next;
    }
}

class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null) return null;
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenhead = head.next;
        while(even != null && even.next != null){
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        odd.next = evenhead;
        return head;
    }
}

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:  #头结点限定条件
            return head

        odd = head
        even = head.next
        t = even   #！！！记录偶数结点的起始位置
        while even != None and even.next != None:
            #！！！遍历结点，保持奇数偶数结点的相对位置
            odd.next = even.next  #连接奇数结点
            odd = odd.next        #调整位置
            even.next = odd.next  #连接偶数结点
            even = even.next      #调整位置
        odd.next = t              #整合链表
        return head

原文地址：https://www.cnblogs.com/yawenw/p/12915691.html

May LeetCoding Challenge16 之 链表重组

May LeetCoding Challenge16 之链表重组