86. 分隔链表

本来想用快慢指针做的，就像面试题02.04，交换节点的值，但是这样不能保证每个结点的初始相对位置，于是就只能构造两个链表large和small，分别存放大于x的节点和小于x的节点，再把两个链表拼接起来。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* partition(struct ListNode* head, int x){
    //快慢指针，交换顺序
    if(head==NULL || head->next==NULL) return head;
    struct ListNode* large=(struct ListNode*)malloc(sizeof(struct ListNode));
    large->next=NULL; // 如果不加会报错，runtime error: member access within misaligned address
    struct ListNode* headl=large;
    struct ListNode* small=(struct ListNode*)malloc(sizeof(struct ListNode));
    small->next=NULL;
    struct ListNode* heads=small;
    while(head){
        if(head->val <x){
            small->next=head;
            small=head;
        }
        else{
            large->next=head;
            large=head;
        }
        head=head->next;
    }
    small->next=headl->next;
    large->next=NULL;
    return heads->next;
}