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LeetCode86 Partition List

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.(Medium)

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

分析:

把链表归并,其思路就是先开两个链表,把小于x的值接在链表left后面,大于x的值接在链表right后面;

然后把链表left的尾部与链表right的头部接在一起。

注意:链表right的尾部next赋值为nullptr。

代码:

 1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* partition(ListNode* head, int x) {12         ListNode dummy1(0);13         ListNode dummy2(0);14         ListNode* left = &dummy1;15         ListNode* right = &dummy2;16         while (head != nullptr) {17             if (head -> val < x) {18                 left -> next = head;19                 left = left -> next;20             }21             else {22                 right -> next = head;23                 right = right -> next;24             }25             head = head -> next;26         }27         left -> next = dummy2.next;28         right -> next = nullptr;29         return dummy1.next;30     }31 };

 

 

LeetCode86 Partition List