leetcode 086 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

.

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class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode *small=NULL, *greater=NULL, *smallCur=NULL, *greaterCur=NULL;

if(head==NULL) return head;

for(ListNode *cur=head; cur!=NULL;) {
ListNode *next = cur->next;
if(cur->val < x) {
if(small==NULL) {
small=cur;
} else {
smallCur->next=cur;
}
smallCur=cur;
smallCur->next = NULL;
} else {
if(greater==NULL) {
greater=cur;
} else {
greaterCur->next=cur;
}
greaterCur=cur;
greaterCur->next=NULL;
}
cur = next;
}

if(small!=NULL) {
smallCur->next=greater;
} else {
small=greater;
}

return small;
}
};