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leetcode链表--11、partition-list(链表分区)
题目描述
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,Given1->4->3->2->5->2and x = 3,
return1->2->2->4->3->5
解题思路:.本题目是将小于x的放在左侧,大于等于x的放在后面,且相对顺序不变
1)定义两个链表 p存小于x的 q存大于等于x的
2)存完后q->next = NULL
3)p1->next = qhead->next;
1 #include <iostream> 2 #include <malloc.h> 3 using namespace std; 4 struct ListNode { 5 int val; 6 ListNode *next; 7 ListNode(int x) : val(x), next(NULL) {} 8 }; 9 class Solution { 10 public: 11 //定义两个链表 1个存小于x的 一个存大于等于x的 12 //连接两个链表 13 ListNode *partition(ListNode *head, int x) { 14 ListNode *phead = new ListNode(0); 15 ListNode *qhead = new ListNode(0); 16 ListNode *p = phead; 17 ListNode *q = qhead; 18 while(head != NULL) 19 { 20 if(head->val < x) 21 { 22 p->next = head; 23 p = p->next; 24 } 25 else 26 { 27 q->next = head; 28 q = q->next; 29 } 30 head = head->next; 31 } 32 q->next = NULL;//别忘记给q之后的链表断开,否则无限循环 33 p->next = qhead->next; 34 return phead->next; 35 } 36 }; 37 ListNode *CreateList(int n) 38 { 39 ListNode *head; 40 ListNode *p,*pre; 41 int i; 42 head=(ListNode *)malloc(sizeof(ListNode)); 43 head->next=NULL; 44 pre=head; 45 for(i=1;i<=n;i++) 46 { 47 p=(ListNode *)malloc(sizeof(ListNode)); 48 cin>>p->val; 49 pre->next=p; 50 pre=p; 51 } 52 p->next=NULL; 53 54 return head->next; 55 } 56 /*-------------------------输出链表-----------------------------------*/ 57 void PrintList(ListNode *h) 58 { 59 ListNode *p; 60 61 p=h;//不带空的头结点 62 while(p) 63 { 64 cout<<p->val<<" "; 65 p=p->next; 66 cout<<endl; 67 } 68 } 69 int main() 70 { 71 int n1; 72 int x; 73 ListNode *h1; 74 cout<<"输入链表1的结点数目"<<endl; 75 cin>>n1; 76 h1 = CreateList(n1); 77 cout<<"链表1为:"<<endl; 78 PrintList(h1); 79 cout<<"输入x"<<endl; 80 cin>>x; 81 Solution s; 82 h1 = s.partition(h1,x); 83 cout<<"分区后链表1为:"<<endl; 84 PrintList(h1); 85 return 0; 86 }
leetcode链表--11、partition-list(链表分区)
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