标签:des style blog class code c
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
.
不是很麻烦的一道题,首先找到链表中第一个比x大或者等于x的数,它的指针存放在before_p中,那么后面比x小的数都插入到它前面就可以了。在遍历的过程中,还要用指针front_p保存p的前一个指针,方便删除p所指向的节点。开始时候没有用这个指针,后来发现当删除的元素在链表结尾的时候,这个指针是必须的。
一种特殊的情况是需要头插的时候,比如
1. 给定的序列是2->1,给定的x是2,此时1要被头插到2前面去;
2. 给定的序列是2->3->1,x是3的时候1只需插入到2的后面,要注意这两种情况的区别。
利用判断before_p == head && before_p->val >= x,就可以甄别出要头插的情况了。因为在before_p == head时,有两种情况,一种是before_p指向的元素确实比x大或者相等,对应上述的第一种情况,需要头插;另外一种before_p指向的元素比x小,只需要把元素插入到before_p后面就可以了,对应于上述的第二种情况。
WA了一次才发现的,一般好难想到这一点>.<
代码:
1 #include <iostream> 2 #include <stdlib.h> 3 using namespace std; 4 5 struct ListNode { 6 int val; 7 ListNode *next; 8 ListNode(int x) : val(x), next(NULL) {} 9 }; 10 11 class Solution { 12 public: 13 ListNode *partition(ListNode *head, int x) { 14 ListNode* p = head; 15 ListNode* before_p = head; 16 ListNode* front_p; 17 18 //找到链表中第一个大于等于x的数,以后比x小的数都插在它后面 19 while(p!= NULL && p->val < x){ 20 before_p = p; 21 p = p->next; 22 } 23 24 if(p !=NULL) 25 { 26 while(p != NULL){ 27 //找到一个比x小的数,要挪动它 28 if(p->val < x){ 29 int value = p->val; 30 //删除比x的小的数的节点 31 if(p->next != NULL) 32 { 33 ListNode* temp = p->next; 34 p->val = temp->val; 35 p->next = temp->next; 36 free(temp); 37 } 38 else{ 39 front_p->next = NULL; 40 ListNode* temp = p; 41 p = NULL; 42 free(temp); 43 } 44 //删除掉的节点插入到before_p后面 45 ListNode* new_node= (struct ListNode*)malloc(sizeof(struct ListNode)); 46 new_node->val = value; 47 48 if(before_p == head && before_p->val >= x){//头插 49 new_node->next = head; 50 head = new_node; 51 before_p = head; 52 } 53 else//不需要头插 54 { 55 new_node->next = before_p->next; 56 before_p->next = new_node; 57 before_p = new_node; 58 } 59 } 60 else{ 61 front_p = p; 62 p = p->next; 63 } 64 } 65 } 66 return head; 67 } 68 }; 69 70 int main(){ 71 struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode)); 72 head->val = 2; 73 struct ListNode*next = (struct ListNode*)malloc(sizeof(struct ListNode)); 74 next->val = 1; 75 head->next = next; 76 next->next = NULL; 77 /*struct ListNode* node3 = (struct ListNode*)malloc(sizeof(struct ListNode)); 78 node3->val = 3; 79 next->next = node3; 80 struct ListNode* node4 = (struct ListNode*)malloc(sizeof(struct ListNode)); 81 node4->val = 2; 82 node3->next = node4; 83 struct ListNode* node5 = (struct ListNode*)malloc(sizeof(struct ListNode)); 84 node5->val = 5; 85 node4->next = node5; 86 struct ListNode* node6 = (struct ListNode*)malloc(sizeof(struct ListNode)); 87 node6->val = 2; 88 node5->next = node6; 89 node6->next = NULL;*/ 90 91 Solution solution; 92 ListNode* h = solution.partition(head,2); 93 cout <<"dfs"<<endl; 94 while(h != NULL) 95 { 96 cout << h->val<<endl; 97 h = h ->next; 98 } 99 }
【leetcode】Partition List,布布扣,bubuko.com
标签:des style blog class code c
原文地址:http://www.cnblogs.com/sunshineatnoon/p/3731728.html