那些让你觉得自己是个傻B的题目集锦(大神的降维打击合集)

一起过来排好队，进来挨打

1.Leetcode tag-LinkList [109.convert sorted list to binary search tree](#109.convert sorted list to binary search tree)

2Leetcode tag-Array[386. Lexicographical Numbers](#386. Lexicographical Numbers)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        return RecursionListToBST(head,NULL);
    }
    TreeNode* RecursionListToBST(ListNode* head,ListNode *tail){
        // linklist is empty
        if(head==tail)
            return NULL;
        // only one node in the tree
        if(head->next==tail){
            TreeNode *root=new TreeNode(head->val);
            return root;
        }
        // search the middle node
        // excllent code segment need memorize.
        ListNode *mid=head;
        ListNode *temp=head;
        while(temp!=tail && temp->next!=tail){
            mid=mid->next;
            temp=temp->next->next;
        }
        TreeNode *root=new TreeNode(mid->val);
        root->left=RecursionListToBST(head,mid);
        root->right=RecursionListToBST(mid->next,tail);
        return root;
    }
};

// 寻找链表中点这个真的是棒

 ListNode *mid=head;
 ListNode *temp=head;
 while(temp!=tail && temp->next!=tail){
         mid=mid->next;
         temp=temp->next->next;
  }

386. Lexicographical Numbers 按字典进行排序

这种解决问题的思考方式

class Solution {
public:
    vector<int> lexicalOrder(int n) {
        vector<int> res(n);
        int cur=1;
        for(int i=0;i<n;i++)    //进行循环遍历
        {
            res[i]=cur;
            if(cur*10<=n)
                cur*=10;     // 进行倍数更新
            else
            {
                if(cur>=n)
                    cur/=10;
                cur+=1;      //保持自增
                while(cur%10==0)
                    cur/=10;
            }
        }
        return res;
    }
};