LeetCode:Partition List
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Partition List
Total Accepted: 63877 Total
Submissions: 217363 Difficulty: Medium
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
.
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思路:
用两个链表分别链接小于x和大于x的结点,然后将两个链表合并。
code:
/** * 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 dummy1(0),dummy2(0); // 两个伪头结点 ListNode *p1 = &dummy1,*p2 = &dummy2; // 两个移动指针 while(head){ if(head->val < x){ p1->next = head; p1 = p1->next; }else{ p2->next = head; p2 = p2->next; } head = head->next; } p2->next = NULL; p1->next = dummy2.next; return dummy1.next; } };
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