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

.

关于指针的操作！需要熟练，多做久好了。

/**
* 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 || head->next == NULL) return head;
ListNode *newHead = NULL;
ListNode *smaller = NULL;
ListNode *largehead = NULL;
ListNode *larger = NULL;
while(head != NULL) {
if(head->val < x) {
if(smaller == NULL) {
smaller = head;
newHead = head;
}
else {
smaller->next = head;
smaller = smaller->next;
}
}
else {
if(larger == NULL) {
larger = head;
largehead = head;
}
else {
larger->next = head;
larger = larger->next;
}
}
head = head->next;
}

if(larger != NULL) {
larger->next = NULL;
}

if(smaller != NULL) {
smaller->next = largehead;
return newHead;
}

return largehead;
}
};

Partition List