Partition List

描述

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.

分析

无

代码

Java

// Partition List
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode left_dummy = new ListNode(-1); // 头结点
        ListNode right_dummy = new ListNode(-1); // 头结点

        ListNode left_cur = left_dummy;
        ListNode right_cur = right_dummy;

        for (ListNode cur = head; cur != null; cur = cur.next) {
            if (cur.val < x) {
                left_cur.next = cur;
                left_cur = cur;
            } else {
                right_cur.next = cur;
                right_cur = cur;
            }
        }

        left_cur.next = right_dummy.next;
        right_cur.next = null;

        return left_dummy.next;
    }
};

C++

// Partition List
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode left_dummy(-1); // 头结点
        ListNode right_dummy(-1); // 头结点

        auto left_cur = &left_dummy;
        auto right_cur = &right_dummy;

        for (ListNode *cur = head; cur != nullptr; cur = cur->next) {
            if (cur->val < x) {
                left_cur->next = cur;
                left_cur = cur;
            } else {
                right_cur->next = cur;
                right_cur = cur;
            }
        }

        left_cur->next = right_dummy.next;
        right_cur->next = nullptr;

        return left_dummy.next;
    }
};

Python

# Partition List
# 时间复杂度O(n)，空间复杂度O(1)
class Solution:
    def partition(self, head, x):
        left_dummy = ListNode(-1) # 头结点
        right_dummy = ListNode(-1) # 头结点

        left_cur = left_dummy
        right_cur = right_dummy
        cur = head
        while cur is not None:
            if cur.val < x:
                left_cur.next = cur
                left_cur = cur
            else:
                right_cur.next = cur
                right_cur = cur
            cur = cur.next

        left_cur.next = right_dummy.next
        right_cur.next = None

        return left_dummy.next