标签：accept container vector min 新建 types comment notice tac
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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | /** * 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) { /** we can use two vector<int> store different types of numbers, n1 record the number smaller than x, n2 record * the numbers no-smaller than x */ if(head == NULL || head->next == NULL) return head; vector<int> n1, n2; ListNode* p = head; while(p != NULL){ p->val < x ? n1.push_back(p->val) : n2.push_back(p->val); p = p->next; } p = head; for(auto n: n1){ p->val = n; p = p->next; } for(auto n: n2){ p->val = n; p = p->next; } return head; }}; |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 | /** * 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) { if(head == NULL || head->next == NULL) return head; ListNode dummy1(0), dummy2(0); ListNode* p1 = &dummy1, * p2 = &dummy2; while(head != NULL){ // using reference to reduce the code lines ListNode* & ref = head->val < x ? p1 : p2; ref->next = head; ref = ref->next; /** if(head->val < x){ p1->next = head; p1 = p1->next; } else{ p2->next = head; p2 = p2->next; } */ head = head->next; } p1->next = dummy2.next; // notice that this step makes sure that there is no loop in the new list p2->next = NULL; return dummy1.next; }}; |
标签：accept container vector min 新建 types comment notice tac
原文地址：http://www.cnblogs.com/zhxshseu/p/2fce572ddcac587809a8981aa94095c5.html
