LeetCode:Partition List

Partition List

   

Total Accepted: 63877 Total Submissions: 217363 Difficulty: Medium

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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 Linked List Two Pointers

思路：

用两个链表分别链接小于x和大于x的结点，然后将两个链表合并。

code:

/**
 * 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) {
        
        ListNode dummy1(0),dummy2(0); // 两个伪头结点
        ListNode *p1 = &dummy1,*p2 = &dummy2; // 两个移动指针
        
        while(head){
            if(head->val < x){
                p1->next = head;
                p1 = p1->next;
            }else{
                p2->next = head;
                p2 = p2->next;
            }
            head = head->next;
        }
        p2->next = NULL;
        p1->next = dummy2.next;
        return dummy1.next;
    }
};