【Leetcode】【Medium】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

.

解题思路：

1、先条件：输入链表不为空；

2、表头可能改变，因此需要新建一个结点指向表头，或使用二维指针；

3、不变参数：

　　curNode永远指向待操作结点的前驱（方便删减）

　　small_end永远指向已经分割的所有小结点中，最后一个结点

　　big_begin永远指向已经分割的所有大结点中，第一个结点

　　small_end->next = big_begin;

4、curNode->next为NULL时，循环结束。当发现小结点时，插入small_end和big_begin中间；

代码：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
if (head == NULL)
return head;
ListNode* prehead = new ListNode(0);
prehead->next = head;
ListNode* curNode = prehead;
ListNode* small_end = NULL;
ListNode* big_begin = NULL;

while (curNode->next && curNode->next->val < x)
curNode = curNode->next;

small_end = curNode;
big_begin = small_end->next;

while (curNode->next) {
if (curNode->next->val < x) {
small_end->next = curNode->next;
small_end = small_end->next;
curNode->next = curNode->next->next;
small_end->next = big_begin;
} else {
curNode = curNode->next;
}
}

head = prehead->next;
delete prehead;
return head;
}
};