leetcode 876. Middle of the Linked List

不容易出错的写法，慢

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        if(!head || !head->next) {
            return head;
        }
        ListNode *single = head, *double_ = head;
        int cnt = 0;
        while(double_) {
            cnt++;
            double_ = double_->next;
            if((cnt & 1) == 0) {
                single = single->next;
            }
        }
        return single;
    }
};

这里用cnt计数会拖慢速度，可以考虑优化边界条件判断，取消计数器

1ms写法

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        ListNode *single = head, *double_ = head;
        while(double_ && double_->next) {
            double_ = double_->next->next;
            single = single->next;
        }
        return single;
    }
};

只要保证double_和double_->next都不空，就不会访问到nullptr的成员，就不会出现runtime error