题解 | #链表的奇偶重排#

/**
 * struct ListNode {
 *    int val;
 *    struct ListNode *next;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @return ListNode类
 */
#include<stdlib.h>
#include<stdio.h>
struct ListNode* oddEvenList(struct ListNode* head ) {
    // write code here
    //计算节点个数
    struct ListNode *p1 = head;
    int count = 0;
    while(p1 != NULL)
    {
        p1 = p1->next;
        count++;
    }
    if(count == 2|| count == 1||count == 0) return head;
    int old = count/2,even = (count+1)/2;
    struct ListNode *p2;
    struct ListNode *newhead = (struct ListNode*)malloc(sizeof(struct ListNode));
    p2 = newhead;
    p1 = head;
    for(int i = 1;i<even;i++)
    {
        p2->val = p1->val;
        p1 = p1->next->next;
        p2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        p2 = p2->next;
    }
    p2->val = p1->val;
    //奇数完
    p1 = head->next;
    p2->next = (struct ListNode *)malloc(sizeof(struct ListNode));
    p2 = p2->next;
    for(int i = 1;i<old;i++)
    {
        p2->val = p1->val;
        p1 = p1->next->next;
        p2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        p2 = p2->next;
    }
    p2->val = p1->val;
    p2->next = NULL;
    //偶数完
    return newhead;
}