LeetCode: Partition List

一次过，链表题无难度

 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         // Start typing your C/C++ solution below
13         // DO NOT write int main() function
14         ListNode *less, *larger, *p, *q, *ret;
15         less = larger = p = q = ret = NULL;
16         while (head) {
17             if (head->val < x) {
18                 if (!less) {
19                     less = new ListNode(head->val);
20                     p = less;
21                 }
22                 else {
23                     p->next = new ListNode(head->val);
24                     p = p->next;
25                 }
26             }
27             else {
28                 if (!larger) {
29                     larger = new ListNode(head->val);
30                     q = larger;
31                 }
32                 else {
33                     q->next = new ListNode(head->val);
34                     q = q->next;
35                 }
36             }
37             head = head->next;
38         }
39         if (!less) ret = larger;
40         if (!larger) ret = less;
41         if (less && larger) {
42             p->next = larger;
43             ret = less;
44         }
45         return ret;
46     }
47 };

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public int val;
 5  *     public ListNode next;
 6  *     public ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode Partition(ListNode head, int x) {
11         ListNode less = null, larger = null, p = null, q = null, ans = null;
12         while (head != null) {
13             if (head.val < x) {
14                 if (less == null) {
15                     less = new ListNode(head.val);
16                     p = less;
17                 }
18                 else {
19                     p.next = new ListNode(head.val);
20                     p = p.next;
21                 }
22             }
23             else {
24                 if (larger == null) {
25                     larger = new ListNode(head.val);
26                     q = larger;
27                 }
28                 else {
29                     q.next = new ListNode(head.val);
30                     q = q.next;
31                 }
32             }
33             head = head.next;
34         }
35         if (less == null) ans = larger;
36         if (larger == null) ans = less;
37         if (less != null && larger != null) {
38             p.next = larger;
39             ans = less;
40         }
41         return ans;
42     }
43 }

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posted @ 2013-04-19 03:36 ying_vincent 阅读(150) 评论(0) 收藏举报

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