Leetcode 328 Odd Even Linked List 链表

将链表节点序号（不是值）是偶数的放到链表后面, 如

Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

我首先统计了下链表的大小cnt,同时求出链表尾端end，

然后直接将每个链表节点序号是奇数的点后面的节点放到end后面去，

同时更新end,这样更新cnt/2次。

注意点：当链表长度小于 3的时候不需要做这样的操作。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(!head) return head;
        ListNode* last = head;
        int cnt = 1;
        for(; last->next; last = last->next, ++cnt);
        if(cnt < 3) return head;
        ListNode* now = head;
        ListNode* end = last;
            
        for(int i = 0; i< cnt/2; now = now->next, ++i){
            
            ListNode* next = now->next;
            now->next = next->next;
                
            end->next = next;
            next->next = NULL;
                
            end = next;
            
        }
        return head;
    }
};

这里在附上一段测试代码

int main(){
    int n;
    while(scanf("%d",&n)){
        ListNode* head = new ListNode(1);
        ListNode* node= head;
        for(int i = 2; i<=n; ++i ){
            ListNode* now = new ListNode(i);
            node->next = now;
            node = now;
        }
        Solution s;
        s.oddEvenList(head);
        for(node = head; node; node = node ->next){
            printf("%d ",node->val);
        }
        puts("");
        for(node = head; node; ){
            ListNode* now = node;
            node = node ->next;
            delete now;
        }
    }
}