leetcode: Partition List

最新推荐文章于 2018-10-12 09:53:13 发布

原创最新推荐文章于 2018-10-12 09:53:13 发布 · 493 阅读

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本文详细阐述了如何通过链表节点值将链表分为两部分，一部分包含小于给定值x的所有节点，另一部分包含大于等于x的所有节点，同时保持原始相对顺序不变。

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Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal tox.

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.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
        
        ListNode* pSHead = NULL;
        ListNode* pBHead = NULL;
        
        ListNode* pSPre  = NULL;
        ListNode* pBPre  = NULL;
        
        ListNode* pCur = head;
        
        while (pCur)
        {
            if (pCur->val < x)
            {
                if (pSHead == NULL)
                {
                    pSHead = pCur;
                    pSPre  = pCur;
                }
                else
                {
                    pSPre->next = pCur;
                    pSPre = pCur;
                }
            }
            else
            {
                if (pBHead == NULL)
                {
                    pBHead = pCur;
                    pBPre  = pCur;
                }
                else
                {
                    pBPre->next = pCur;
                    pBPre = pCur;
                }
            }
            pCur = pCur->next;
        }
        
        if (pSHead == NULL)
            return pBHead;
            
        if (pBHead == NULL)
            return pSHead;
            
        pBPre->next = NULL;    
        pSPre->next = pBHead;
        
        return pSHead;
    }
};