pay for ; the difference ; their improve without write stop parents instead of a way seldom respect

Posted 2023-04-21

How many times have you felt that your parents don't understand you?Parents!They are like people from another planet!There is____of bridging the gap.
If you want to _____your relationship with your parents,try listening to them,treating then just like your friends.____always saying "You don't understand me ,"____and think .Do you ener try and understand them?Parents are under a lot of stress ,too. When you are being worried about your upcoming math exam, they are being worried about their bosses in the office, and just now they are going to ____ your expensive shoes.Your mom mkay ___have a chance to go out and enjoyed herself.
Parents are humans ,too.Once you step into their shoes and try to look at things from ____ point of view,you will find a new ____for them. And you will find you can get your own way ____quarrels.
So try to ask your mom or dad,"How did your day go today?" or "Is there anything I can do to help you around the house?" when you come back from school.You 'll see ____it make to the atmosphere(气氛） at home.

参考技术A 词组短语的东西 量比5较大e 不x复制过来了j 给几l个m连接 初中2必备短语 初三w常用短语动词分2类归纳 。;。。 初中4英语常用介0词短语 。;。。 另外建议生活中2购买 生活中0好的也g很多的 可以4去书5店逛逛 网上b的东西看过眼就忘了z 下i次比7较难找nb咯ΗpЬkけ伐uz⑧wˇe 参考技术B 1. a way 2.improve 3.instead of 4. stop 5.pay for 6.seldom 7.the difference
8. respect 9.without 10.本回答被提问者采纳 参考技术C 不会

Add Digits, Maximum Depth of BinaryTree, Search for a Range, Single Number,Find the Difference

最近做的题记录下。

258. Add Digits

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

 1 int addDigits(int num) {
 2         char strint[12] = {0};
 3         sprintf(strint, "%d", num);
 4         int i=0,a = 0;
 5         while(strint[i+1]!= ‘\0‘)
 6         {
 7            a = (strint[i]-‘0‘)+(strint[i+1]-‘0‘);
 8            if(a>9)
 9            {
10                 a = a%10+a/10;
11            }
12            strint[++i] = a+‘0‘;
13         }
14         return strint[i]-‘0‘;
15 }

上边这是O(n)的复杂度。网上还有O(1)的复杂度的：return (num - 1) % 9 + 1;

104. Maximum Depth of Binary Tree

求二叉树的深度，可以用递归也可以用循环，如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     struct TreeNode *left;
 6  *     struct TreeNode *right;
 7  * };
 8  */
 9 int maxDepth(struct TreeNode* root) {
10     int left = 0,right = 0;
11     if (root == NULL) return 0;
12 
13     left = maxDepth(root->left);
14     right = maxDepth(root->right);
15 
16     return left>right? left+1:right+1;
17 }

******************************************************************************************************************

用队列存结点，记录每一层的结点数levelCount，每出一个结点levelCount--，如果减为0说明要进入下一层，然后depth++，同时队列此时的结点数就是下一层的结点数。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int maxDepth(TreeNode* root) {
13         if (root == NULL) return 0;
14 
15         //深度和每一层节点数
16         int depth = 0,levelCount = 1;
17         //队列保存每一层的节点数
18         queue<TreeNode*> node;
19 
20         node.push(root);
21         while(!node.empty())
22         {
23             //依次遍历每一个结点
24             TreeNode *p = node.front();
25             node.pop();
26             levelCount--;
27 
28             if(p->left) node.push(p->left);
29             if(p->right) node.push(p->right);
30 
31             if (levelCount == 0)
32             {
33                 //保存下一层节点数，深度加1
34                 depth++;
35                 levelCount = node.size();
36             }
37         }
38         return depth;
39     }
40 };

34. Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm‘s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,

Given [5, 7, 7, 8, 8, 10] and target value 8,

return [3, 4]

 1 vector<int> searchRange(vector<int>& nums, int target) {
 2         int size = nums.size();
 3         int l=0,r=size-1;
 4         while(l<=r)
 5         {
 6             int mid = (l+r)/2;
 7             if (nums[mid] >= target)
 8             {
 9                 r = mid-1;
10             }
11             else if(nums[mid] < target)
12             {
13                 l = mid+1;
14             }
15         }
16         int left = l;
17         l = 0;
18         r = size-1;
19         while(l<=r)
20         {
21             int mid = (l+r)/2;
22             if (nums[mid] <= target)
23             {
24                 l = mid+1;
25             }
26             else if(nums[mid] > target)
27             {
28                 r = mid-1;
29             }
30         }
31         int right = r;
32         vector<int> v;
33         v.push_back(left);
34         v.push_back(right);
35         if(nums[left] != target || nums[right] != target)
36         {
37             v[0] = -1;
38             v[1] = -1;
39         }
40         return v;
41     }

这个题就是要把握好边界，比如找左边界时 =号要给右边界的判断，找右边界则相反。这样才能保证不遗漏

136. Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

直接用异或

 1 int singleNumber(vector<int>& nums) {
 2         int result=0;
 3         int len = nums.size();
 4         if (len <=0) return 0;
 5         for(int i=0;i<len;i++)
 6         {
 7             result ^= nums[i];
 8         }
 9         return result;
10     }

389. Find the Difference

Given two strings s and t which consist of only lowercase letters.

String t is generated by random shuffling string s and then add one more letter at a random position.

Find the letter that was added in t

 

这种方法挺慢，我提交显示16ms，没有那种用char数组来模拟hash快。

 1 char findTheDifference(string s, string t) {
 2         map<char, int> sumChar;
 3         char res;
 4         for(auto ch: s) sumChar[ch]++;
 5         for(auto ch: t)
 6         {
 7            if(--sumChar[ch]<0)
 8                res = ch; 
 9         }    
10         return res;
11     }

这道题也可以用异或 因为相同的会异或掉，剩下一个就是多余的char。