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.

/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(head == NULL) return head;
ListNode *pre , *firstLarger, * preCur, * cur;
pre = NULL; firstLarger = head;
while(firstLarger  && firstLarger->val < x){
pre = firstLarger;
firstLarger = firstLarger ->next;
}
if(firstLarger == NULL) return head;

preCur = firstLarger;
cur = preCur->next;

while(cur != NULL){
if( cur->val >= x){
preCur = cur;
cur = cur->next;
continue;
}

preCur->next = cur->next;
cur->next = firstLarger;
if(pre == NULL){
pre = cur;
head = cur;
}else{
pre->next = cur;
pre = cur;
}
cur = preCur ->next;
}

return head;
}
};