[LeetCode April Challange] Day 14 - Partition List

Given the head of 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 1:

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

Input: head = [2,1], x = 2
Output: [1,2]

Constraints:

• The number of nodes in the list is in the range [0, 200].
• -100 <= Node.val <= 100
• -200 <= x <= 200

---

Solution

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode *lt_dummy_head = new ListNode();
        ListNode *egt_dummy_head = new ListNode();
        ListNode *lt = lt_dummy_head;
        ListNode *egt = egt_dummy_head;
        for (ListNode *curr=head; curr; curr=curr->next) {
            if (curr->val < x) {
                lt->next = curr;
                lt = lt->next;
            } else {
                egt->next = curr;
                egt = egt->next;
            }
        }
        egt->next = nullptr;
        lt->next = egt_dummy_head->next;
        return lt_dummy_head->next;
    }
};

從頭到尾掃一遍，用兩個指標記錄 小於x & 大於等於x 的 node 順序。
掃完後，將 lt 與 egt 串起來，egt 下一個設為 nullptr，完工。