目錄

1.加一

2.二進制求和

3.x的平方根

4.爬樓梯

5.顏色分類

6.二叉樹的中序遍歷

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1.加一

https://leetcode.cn/problems/plus-one/

class Solution {
public:vector<int> plusOne(vector<int>& digits) {int n = digits.size();for(int i = n -1;i>=0;--i){if(digits[i] != 9){++digits[i];for(int j = i + 1;j<n;++j){digits[j] = 0;}return digits;}}//數組全部為9vector<int> ans(n+1);ans[0] = 1;return ans;}
};

2.二進制求和

https://leetcode.cn/problems/add-binary/description/

class Solution {
public:string addBinary(string a, string b) {int i = a.length() - 1;int j = b.length() - 1;int carry = 0;string builder;//循環相加兩個字符串相同長度的低位數部分while (i >= 0 && j >= 0) {int sum = carry;std::cout<<"a---"<<a.at(i)-'0'<<std::endl;std::cout<<"b---"<<b.at(j)-'0'<<std::endl;sum += a.at(i--) - '0';sum += b.at(j--) - '0';carry = sum / 2; //是否進位builder+=to_string((sum % 2));}// 如果 a 還沒遍歷完成（a串比b串長），則繼續遍歷添加 a 的剩余部分while (i >= 0) {int sum = carry + a.at(i--) - '0';carry = sum / 2;builder+=to_string(sum % 2);}// 如果 b 還沒遍歷完成（b串比a串長），則繼續遍歷添加 b 的剩余部分while (j >= 0) {int sum = carry + b.at(j--) - '0';carry = sum / 2;builder+=to_string(sum % 2);}//如果 carry 不等于0 還有個進位數沒加進去，需要補充if (carry == 1) {builder+=to_string(carry);}//反轉字符串獲得正常結果reverse(builder.begin(),builder.end());return builder;}
};

3.x的平方根

https://leetcode.cn/problems/sqrtx/

class Solution {
public:int mySqrt(int x) {//使用庫中函數int sum = sqrt(x);return sum;}
};

4.爬樓梯

https://leetcode.cn/problems/climbing-stairs/description/

class Solution {
public:int climbStairs(int n) {//f(x)=f(x-1)+f(x-2);int p = 0, q = 0, r = 1;for (int i = 1; i <= n; ++i) {p = q; q = r; r = p + q;}return r;}
};

5.顏色分類

https://leetcode.cn/problems/sort-colors/description/

class Solution {
public:void sortColors(vector<int>& nums) {//sort(nums.begin(),nums.end());//p0交換0,p1交換1int p0 = 0,p1 = 0;for(int i = 0;i < nums.size();++i){if(nums[i] == 1){swap(nums[i],nums[p1]);++p1;}else if(nums[i] == 0){swap(nums[i],nums[p0]);if(p0 < p1){swap(nums[i],nums[p1]);}++p0;++p1;}}}
};

6.二叉樹的中序遍歷

https://leetcode.cn/problems/binary-tree-inorder-traversal/description/

class Solution {
public:void inorder(TreeNode* root,vector<int>& res){if(root == NULL)return;inorder(root->left,res);res.push_back(root->val);inorder(root->right,res);}vector<int> inorderTraversal(TreeNode* root) {vector<int> res;inorder(root,res);return res;}
};