LeetCode --- 86. Partition List

86. Partition List

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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

Solution

思路

使用两个链表，before_head保存比x值小的结点，after_head保存比x大的结点，最后将两个链表进行拼接。
Language: C++

/**
 * 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) return head;
        ListNode *before_head = new ListNode(0);
        ListNode *after_head = new ListNode(0);
        ListNode *before = before_head, *after = after_head;
        while (head){
            if (head->val < x){
                before->next = head;
                before = before->next;
            }
            else{
                after->next = head;
                after = after->next;
            }
            head = head->next;
        }
        after->next = NULL;
        before->next = after_head->next;
        return before_head->next;
    }
};