Partition List

最新推荐文章于 2021-04-06 09:37:45 发布

原创最新推荐文章于 2021-04-06 09:37:45 发布 · 843 阅读

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本文介绍了一种链表分区算法，该算法将链表中的节点按给定值x分为两部分，所有小于x的节点置于大于等于x的节点之前，并保持各部分内部节点的原有顺序。提供了两种实现方案，包括具体的C++代码示例。

问题：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.

思路：遍历一遍。将这一个链表拆成两条链表，一个里面全是小于x的，另一个里面是大于等于x的。新链表都在尾部插入新结点，可以保证相对顺序不变。注意特殊情况：可能整个原链表中的结点都是小于（或者大于）x的。

代码一：

/**
 * 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) {
        if(head == NULL)
            return NULL;
        ListNode *big_head = NULL, *small_head = NULL;
        ListNode *big_end, *small_end;
        
        while(head != NULL)
        {
            if(head->val >= x)
                add_big_list(big_head, big_end, head);
            else
                add_small_list(small_head, small_end, head);
            head = head->next;
        }
        if(big_head != NULL && small_head != NULL)
        {
            small_end->next = big_head;
            big_end->next = NULL;
            return small_head;
        }
        if(big_head == NULL)
        {
            small_end->next = NULL;
            return small_head;
        }
        else
        {
            big_end->next = NULL;
            return big_head;
        }
    }
    
    void add_big_list(ListNode* &head, ListNode* &end, ListNode *p)
    {
        if(head == NULL)
        {
            head = p;
            end = p;
            return;
        }
        end->next = p;
        end = p;
    }
    
    void add_small_list(ListNode* &head, ListNode* &end, ListNode *p)
    {
        if(head == NULL)
        {
            head = p;
            end = p;
            return;
        }
        end->next = p;
        end = p;
    }
};

代码二：优化。一、为两个新链表建立空头结点来使操作保持一致。二、不创建函数了，直接在对原始链表的遍历循环中做插入操作。

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode small_head(0);
        ListNode big_head(0);
        
        ListNode *p1 = &small_head;
        ListNode *p2 = &big_head;
        
        while(head != NULL)
        {
            if(head->val < x)
            {
                p1->next = head;
                p1 = p1->next;
            }
            else
            {
                p2->next = head;
                p2 = p2->next;
            }
            head = head->next;
        }
        p1->next = big_head.next;
        p2->next = NULL;
        return small_head.next;
    }
};