LeetCode-143-重排链表-C语言

/*
 * 算法思想：
 * 将链表分为两段，其实是构造了两个链表，第一个链表使用的节点为前（len+1）/2个节点；
 * 第二个链表为len-(len+1)/2个节点；第二个链表构造的同时反转该链表；
 * 如链表1,2,3,4,5,6,7; 构造后list1 = 1，2，3，4; list2 = 7,6,5;
 * 再将两个链表构造成一个链表，分别循环取list1,list2中的节点即可。构造后为：[1,7,2,6,3,5,4]
 *
 *
 */

typedef struct ListNode Node;

void print(Node *head){
    return;
    while(head){
        printf("%d->", head->val);
        head = head->next;
    }
    printf("\n");
    return;
}

int get_len(Node *head){
    int i=0;
    while(head){
        head = head->next;
        i++;
    }
    return i;
}

void reorderList(struct ListNode* head){
     Node node1;node1.next=NULL;
    Node *rear1 = &node1;
     Node node2;node2.next = NULL;
    Node *rear2 = &node2;
    int i =0;
    Node *p = head;
    Node *last = NULL;
    int len = get_len(head);
    
    while(p || last){
        if(i<(len+1)/2 ){
            /* add nodes to list1 from rear */
            if(last){
                rear1->next = last;
                rear1 = rear1->next;
                rear1->next = NULL;    
                i++;
            }
        }else{
            /* add nodes to lsit 2 from head */
            last->next = node2.next;
            node2.next = last;
        }
        
        last = p;
        if(p)
            p = p->next;
    }
    
    print(node1.next);
    print(node2.next);
    
    /* add two list */
    Node *p1 = node1.next;
    Node *p2 = node2.next;
    Node node3;
    node3.next = NULL;
    Node *rear3 = &node3;
    Node *last1 = NULL;
    Node *last2 = NULL;
    
    while(p1 || p2 || last1 || last2){
        
        if(last1){
            rear3->next = last1;
            rear3 = rear3->next;
            rear3->next = NULL;
        }
        if(last2){
            rear3->next = last2;
            rear3 = rear3->next;
            rear3->next = NULL;
        }
        
        last1 = p1;
        if(p1)
            p1 = p1->next;
        last2 = p2;
        if(p2)
            p2 = p2->next;
    }
    
    return node3.next;
    
}