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Partition List⚓︎

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Description⚓︎

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⚓︎

See reference (Chinese).

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/**
 * 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) {
        auto leftDummyHead = new ListNode(), rightDummyHead = new ListNode();
        auto leftTail = leftDummyHead, rightTail = rightDummyHead;

        while (head) {
            if (head->val < x) {
                leftTail->next = head;
                leftTail = leftTail->next;
            } else {
                rightTail->next = head;
                rightTail = rightTail->next;
            }
            head = head->next;
        }

        leftTail->next = rightDummyHead->next;
        rightTail->next = nullptr;

        return leftDummyHead->next;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).

Note: The evaluation program will traverse the result linked list once, if rightTail->next = nullptr; is not written here, the end of the linked list may point to a node in the original linked list, so that traversing the result linked list may result in an endless loop.