leetcode_partition_list

发表于

难度：Medium

解题思路：画图

value = 3

l_dummy->null        1 -> 4 -> 3 -> 2 -> 5 -> null    
g_dummy->null

l_dummy--------> 1 -> 4 -> 3 -> 2 -> 5 -> null
g_dummy->null


l_dummy--------> 1 -> 4 -> 3 -> 2 -> 5 -> null
g_dummy---------------^

l_dummy--------> 1 -> 4 -> 3 -> 2 -> 5 -> null
g_dummy---------------^

                 -------->------
l_dummy--------> 1    4 -> 3 -> 2 -> 5 -> null
g_dummy---------------^

                 -------->------
l_dummy--------> 1    4 -> 3    2 -> 5 -> null
g_dummy---------------^    ----------^

代码如下：

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/**
 * 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) {
        ListNode *l_dummy = new ListNode(-1), *l_current = l_dummy;
        ListNode *g_dummy = new ListNode(-2), *g_current = g_dummy;
        ListNode *current = head;
        while(current != NULL)
        {
            // cout<<"current:"<<current->val<<endl;
            if(current->val < x)
            {
                l_current->next = current;
                l_current = l_current->next;
            }
            else
            {
                g_current->next = current;
                g_current = g_current->next;
            }
            current = current->next;
        }
        // cout<<"l_dummy->next"<<l_dummy->next->val<<endl;
        // cout<<"g_dummy->next"<<g_dummy->next->val<<endl;
        if(l_current == l_dummy && g_current == g_dummy)
        {
            return l_current->next;
        }
        else if(l_current == l_dummy)
        {
            return g_dummy->next;
        }
        else if(g_current == g_dummy)
        {
            return l_dummy->next;
        }
        else
        {
            l_current->next = g_dummy->next;
            g_current->next = NULL;
            return l_dummy->next;
        }
            
    }
};

运行结果：6ms，超过18.98%