Leetcode_partition-list

最新推荐文章于 2020-08-02 18:53:11 发布

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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指定值x进行划分，使得所有小于x的节点位于大于等于x的节点之前，并保持原有相对顺序不变。提供了三种不同的实现方法，包括详细的代码示例。

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地址： http://oj.leetcode.com/problems/partition-list/

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.

思路：把原来链表拆成两个单链表，一个是小于x的，一个是大于等于x的，最后将第一个链表的尾节点的next域指向第二个节点的头节点，来拼成一个节点。

写了两份，思路都是一样的，第一个繁杂一点，第二个简明一点。

参考代码1：

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(!head)
            return NULL;
        ListNode *new_head = head, *p = head, *bigger_head = NULL, *pre=NULL, *bigger_cur=NULL;
        while(new_head && new_head->val>=x)
            new_head = new_head->next;
        //all elems >= x
        if(!new_head)
            return head;
        while(p)
        {
            if(p->val<x)
            {
                pre = p;
                p = p->next;
            }
            else
            {
                if(!bigger_head)
                    bigger_cur = bigger_head = p;
                else
                {
                    bigger_cur->next = p;
                    bigger_cur = bigger_cur->next;
                }
                p = p->next;
                while(p && p->val >= x)
                {
                    bigger_cur = p;
                    p = p->next;
                }
                if(p)
                {
                    if(!pre)
                        pre = p;
                    else
                    {
                        pre->next = p;
                        pre = p;
                    }
                    p = p->next;
                }
            }
        }
        if(bigger_cur)
            bigger_cur->next = NULL;
        if(pre)
            pre->next = bigger_head;
        return new_head;
    }
};

参考代码2:

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(!head)
            return NULL;
        ListNode *newHead = NULL, *newTail=NULL, *biggerHead = NULL, *q = NULL, *p = head;
        while(p)
        {
            if(p->val<x)
            {
                if(!newHead)
                    newTail = newHead = p;
                else
                    newTail = newTail->next = p;
                p = p->next;
            }
            else
            {
                if(!biggerHead)
                    q = biggerHead = p;
                else
                    q = q->next = p;
                p = p->next;
                while(p && p->val>=x)
                {
                    q = q->next = p;
                    p = p->next;
                }
            }
        }
        if(!newHead || !biggerHead)
            return head;
        newTail->next = biggerHead;
        q->next = NULL;
        return newHead;
    }
};

//Solution 3:

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(!head)
            return NULL;
        ListNode *cur = head, *left = NULL, *right = NULL, *curLeft = NULL, *curRight = NULL;
        while(cur) {
            if(cur->val < x) {
                if(left)
                    curLeft = curLeft->next = cur;
                else
                    left = curLeft = cur;
            } else {
                if(right)
                    curRight = curRight->next = cur;
                else
                    right = curRight = cur;
            }
            cur = cur->next;
        }
        if(left) {
            curLeft->next = right;
            if(right)
                curRight->next = NULL;
            return left;
        } else {
            curRight->next = NULL;
            return right;
        }
    }
};