leetcode-328-Odd-Even-Linked-List

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描述

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example:

Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

分析

也是常规题了，画个草图，注意边界情况。

解决方案1(C++)

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(head && head->next && head->next->next){
            ListNode *odd = head;
            ListNode *even = head->next;
            ListNode *tmp = even;
            while(even || odd) {
                odd->next = odd->next->next;
                even->next = even->next->next;
                
                odd = odd->next;
                even = even->next;
                if(odd->next == NULL || even->next == NULL) break;
            }
            odd->next = tmp;
        }
        return head;
    }
};

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