LeetCode - 分隔鏈表

題目描述

給定一個鏈表和一個特定值 x，對鏈表進行分隔，使得所有小於 x 的節點都在大於或等於 x 的節點之前。

你應當保留兩個分區中每個節點的初始相對位置。

示例:

輸入: head = 1->4->3->2->5->2, x = 3
輸出: 1->2->2->4->3->5

我的思路

藉助兩個輔助鏈表實現即可，然後合併兩個鏈表。代碼如下：

/**
 * 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 == NULL) return head;
        ListNode* first = new ListNode(0);
        ListNode* last = new ListNode(0);
        ListNode* cur = first;
        ListNode* cur2 = last;
        while(head != NULL){
            if(head->val < x){
                first->next = new ListNode(0);
                first->next->val = head->val;
                first = first->next;
            }
            else if(head->val >= x){
                last->next = new ListNode(0);
                last->next->val = head->val;
                last = last->next;
            }
            head = head->next;
        }
       if(first != NULL)first->next = cur2->next;
        return cur->next;
    }
};