Partition List

https://leetcode.com/problems/partition-list/description/

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->null and x = 3, return 1->2->2->4->3->5->null.

Thoughts

把linked list中<x的放前面。两个头不断插入, linked list常见套路。

Code

/*
 * @lc app=leetcode id=86 lang=cpp
 *
 * [86] Partition List
 */
/**
 * 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) {
        auto lh = new ListNode(-1), rh = new ListNode(-1), l = lh, r = rh;
        while (head != nullptr) {
            if (head->val < x) {
                l->next = head;
                l = l->next;
            } else {
                r->next = head;
                r = r->next;
            }
            head = head->next;
        }
        l->next = rh->next;
        r->next = nullptr;
        return lh->next;
    }
};

Analysis

做题耗时: 19min

TC: O(n)