LeetCode解题报告（346) -- 86. Partition List

Problem

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:

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

Example 2:

1 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

Analysis

题目给出一个链表，还有一个指定的值x，要求所有比x大的节点都要放在比x小的节点的后面，而且需要保持原有的相对位置。题目很好理解，实际上就是把原来的链表拆成两条，一条是比x小的节点组成的，另一个条是比x大的节点还有x节点本身组成的。最后再把第二条链表接到第一条链表后面即可。

Solution

无

Code

/**
 * 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* dummy = new ListNode(-1);
        ListNode* dummy2 = new ListNode(-1);
        dummy->next = head;
        ListNode* p = dummy;
        ListNode* p2 = dummy2;
        while (p->next) {
            if (p->next->val < x) {
                p = p->next;
            } else {
                ListNode* temp = p->next;
                p->next = temp->next;
                temp->next = NULL;
                p2->next = temp;
                p2 = p2->next;
            }
        }
        
        p->next = dummy2->next;
        return dummy->next;
    }
};

Summary

这道题目难度不大，属于简单的链表操作。这道题目的分享到这里，感谢你的支持！