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;
    }
};