feat: CPP

Author: lucifer

problems/229.majority-element-ii.md

@@ -71,7 +71,36 @@ https://leetcode-cn.com/problems/majority-element-ii/
 
 ## 代码
 
-JavaScript 代码：
+代码支持：CPP，JS
+
+CPP Code:
+
+```cpp
+class Solution {
+public:
+    vector<int> majorityElement(vector<int>& nums) {
+        int c1 = 0, c2 = 0, v1 = 0, v2 = 1;
+        for (int n : nums) {
+            if (v1 == n) ++c1;
+            else if (v2 == n) ++c2;
+            else if (!c1) v1 = n, ++c1;
+            else if (!c2) v2 = n, ++c2;
+            else --c1, --c2;
+        }
+        c1 = c2 = 0;
+        for (int n : nums) {
+            if (v1 == n) ++c1;
+            if (v2 == n) ++c2;
+        }
+        vector<int> v;
+        if (c1 > nums.size() / 3) v.push_back(v1);
+        if (c2 > nums.size() / 3) v.push_back(v2);
+        return v;
+    }
+};
+```
+
+JS Code：
 
 ```js
 /*

problems/279.perfect-squares.md

@@ -11,7 +11,7 @@ https://leetcode-cn.com/problems/perfect-squares/
 示例 1:
 
 输入: n = 12
-输出: 3 
+输出: 3
 解释: 12 = 4 + 4 + 4.
 示例 2:
 
@@ -31,7 +31,7 @@ https://leetcode-cn.com/problems/perfect-squares/
 - 阿里
 - 百度
 - 字节
-  
+
 ## 思路
 
 直接递归处理即可，但是这种暴力的解法很容易超时。如果你把递归的过程化成一棵树的话（其实就是递归树），
@@ -70,7 +70,7 @@ function d(n, level) {
  * @param {number} n
  * @return {number}
  */
-var numSquares = function(n) {
+var numSquares = function (n) {
   return d(n, 0);
 };
 ```
@@ -98,12 +98,33 @@ for (let i = 1; i <= n; i++) {
 
 ## 代码
 
+代码支持：CPP，JS
+
+CPP Code:
+
+```cpp
+class Solution {
+public:
+    int numSquares(int n) {
+        static vector<int> dp{0};
+        while (dp.size() <= n) {
+            int m = dp.size(), minVal = INT_MAX;
+            for (int i = 1; i * i <= m; ++i) minVal = min(minVal, 1 + dp[m - i * i]);
+            dp.push_back(minVal);
+        }
+        return dp[n];
+    }
+};
+```
+
+JS Code:
+
 ```js
 /**
  * @param {number} n
  * @return {number}
  */
-var numSquares = function(n) {
+var numSquares = function (n) {
   if (n <= 0) {
     return 0;
   }

problems/40.combination-sum-ii.md

@@ -69,7 +69,7 @@ candidates 中的每个数字在每个组合中只能使用一次。
 
 ## 代码
 
-- 语言支持: Javascript，Python3
+- 语言支持: Javascript, Python3, CPP
 
 ```js
 function backtrack(list, tempList, nums, remain, start) {
@@ -142,6 +142,33 @@ class Solution:
                 path.pop()
 ```
 
+CPP Code:
+
+```cpp
+class Solution {
+    vector<vector<int>> ans;
+    void backtrack(vector<int> &A, int target, int start, vector<int> &path) {
+        if (!target) {
+            ans.push_back(path);
+            return;
+        }
+        for (int i = start; i < A.size() && target >= A[i]; ++i) {
+            if (i != start && A[i] == A[i - 1]) continue;
+            path.push_back(A[i]);
+            dfs(A, target - A[i], i + 1, path);
+            path.pop_back();
+        }
+    }
+public:
+    vector<vector<int>> combinationSum2(vector<int>& A, int target) {
+        sort(A.begin(), A.end());
+        vector<int> path;
+        backtrack(A, target, 0, path);
+        return ans;
+    }
+};
+```
+
 ## 相关题目
 
 - [39.combination-sum](./39.combination-sum.md)