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【leetcode每日一题】NO86.Partition List

2015-08-16 09:33 218 查看
题目: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.

解析:先声明两个指针result1和result2,遍历链表,当节点小于x的时候,result1指向该节点,当节点大于或等于x的时候,result2指向该节点。最后再把生成的两个链表进行合并即可。

代码:

/**
 * 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 *result1=new ListNode(0);
        ListNode *result2=new ListNode(0);
        ListNode *temp1=result1;
        ListNode *temp2=result2;
        while(head!=NULL)
        {
            if(head->val<x)
            {
                temp1->next=head;
                temp1=temp1->next;
            }
            else
            {
                temp2->next=head;
                temp2=temp2->next;
            }
            head=head->next;
        }
        temp1->next=result2->next;
        temp2->next=NULL;
        return result1->next;
    }
};
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