57_leetcode_partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

1:特殊情况；2:把链表分成两个子链表；3:最后考虑如果一个子链表为空的情况

    ListNode *partition(ListNode *head,int x)    {        if(head == NULL || head->next == NULL)        {            return head;        }                ListNode* newHead = NULL;        ListNode* index = NULL;        ListNode* bigHead = NULL;        ListNode* bigIndex = NULL;        ListNode* curNode = head;                while(curNode)        {            if(curNode->val >= x)            {                if(bigHead == NULL)                {                    bigHead = curNode;                    bigIndex = bigHead;                }                else                {                    bigIndex->next = curNode;                    bigIndex = bigIndex->next;                }                curNode = curNode->next;            }            else            {                if(newHead == NULL)                {                    newHead = curNode;                    index = newHead;                }                else                {                    index->next = curNode;                    index = index->next;                }               curNode = curNode->next;            }        }                if(index)        {            index->next = bigHead;            if(bigIndex)            {                bigIndex->next = NULL;            }        }        else        {            newHead = bigHead;            bigIndex->next = NULL;        }                return newHead;    }