[update] #3 数组篇添加C++ Code

animation-simulation/剑指offer/1的个数.md

@@ -227,5 +227,37 @@ class Solution {
 
 时间复杂度 : O(logn)  空间复杂度 O(1)
 
+C++ Code:
 
+```C++
+class Solution
+{
+public:
+    int countDigitOne(int n)
+    {
+        //  高位,      低位,    当前位
+        int high = n, low = 0, cur = 0;
+        int count = 0, num = 1;
 
+        //数字是0的时候完全没必要继续计算
+        while (high != 0)
+        {
+            cur = high % 10;
+            high /= 10;
+            //这里我们可以提出 high * num 因为我们发现无论为几，都含有它
+            if (cur == 0)
+                count += (high * num);
+            else if (cur == 1)
+                count += (high * num + 1 + low);
+            else
+                count += ((high + 1) * num);
+            //低位
+            low = cur * num + low;
+            //提前检查剩余数字, 以免溢出
+            if (high != 0)
+                num *= 10;
+        }
+        return count;
+    }
+};
+```

animation-simulation/数组篇/leetcode1052爱生气的书店老板.md

@@ -156,3 +156,49 @@ class Solution {
     }
 }
 ```
+
+C++ Code
+
+```C++
+class Solution
+{
+public:
+    int maxSatisfied(vector<int> &customers, vector<int> &grumpy, int minutes)
+    {
+        for_each(grumpy.begin(), grumpy.end(), [](auto &g){ g = !g; });
+        vector<int> osum(customers.size(), 0);
+
+        //先初始化第一个元素
+        osum[0] = customers[0] * grumpy[0];
+        //计算前缀和, osum是origin sum
+        for (int i = 1; i < osum.size(); i++)
+        {
+            osum[i] = osum[i - 1] + customers[i] * grumpy[i];
+        }
+
+        //计算连续minutes的和
+        vector<int> msum(customers.size() - minutes + 1, 0);
+        for (int i = 0; i < minutes; i++)
+        {
+            msum[0] += customers[i];
+        }
+        for (int i = 1; i < msum.size(); i++)
+        {
+            msum[i] = msum[i - 1] - customers[i - 1] + customers[i + minutes - 1];
+        }
+
+        //分成三段计算
+        int result = 0;
+        for (int i = 0; i < msum.size(); i++)
+        {
+            //左                                         中         右
+            //注意左的边界条件, 可以使用边界测试
+            int sum = ((i - 1 >= 0) ? osum[i - 1] : 0) + msum[i] + osum[osum.size() - 1] - osum[i + minutes - 1];
+            if (sum > result)
+                result = sum;
+        }
+
+        return result;
+    }
+};
+```

animation-simulation/数组篇/leetcode41缺失的第一个正数.md

@@ -232,3 +232,44 @@ class Solution {
     }
 }
 ```
+
+C++ Code
+
+```C++
+class Solution
+{
+public:
+    int firstMissingPositive(vector<int> &nums)
+    {
+        int size = nums.size();
+        //判断范围是否符合要求
+        auto inRange = [](auto s, auto e)
+        {
+            return [s, e](auto &n)
+            {
+                return e >= n && n >= s;
+            };
+        };
+        auto cusInRange = inRange(1, size);
+        //增加数组长度, 便于计算, 不需要再转换
+        nums.push_back(0);
+
+        for (int i = 0; i < size; i++)
+        {
+            //将不在正确位置的元素放到正确位置上
+            while (cusInRange(nums[i]) && nums[i] != i && nums[nums[i]] != nums[i])
+            {
+                swap(nums[i], nums[nums[i]]);
+            }
+        }
+
+        //找出缺失的元素
+        for (int i = 1; i <= size; i++)
+        {
+            if (nums[i] != i)
+                return i;
+        }
+        return size + 1;
+    }
+};
+```

animation-simulation/数组篇/leetcode485最大连续1的个数.md

@@ -177,3 +177,38 @@ class Solution {
     }
 }
 ```
+
+C++ Code
+
+```C++
+class Solution
+{
+public:
+    int findMaxConsecutiveOnes(vector<int> &nums)
+    {
+        int s = 0;
+        int e = 0;
+        int result = 0;
+        int size = nums.size();
+
+        while (s < size && e < size)
+        {
+            while (s < size && nums[s++] == 1)
+            {
+                e = s;
+                while (e < size && nums[e] == 1)
+                {
+                    e++;
+                };
+                //注意需要加1, 可以使用极限条件测试
+                int r = e - s + 1;
+                if (r > result)
+                    result = r;
+                s = e;
+            }
+        }
+
+        return result;
+    }
+};
+```

animation-simulation/数组篇/leetcode560和为K的子数组.md

@@ -186,3 +186,34 @@ class Solution {
     }
 }
 ```
+
+C++ Code
+
+```C++
+class Solution
+{
+public:
+    int subarraySum(vector<int> &nums, int k)
+    {
+        unordered_map<int, int> smp;
+        int sum = 0;
+        //初始化"最外面"的0
+        smp[0] = 1;
+        int result = 0;
+        for(int i = 0; i < nums.size(); i++)
+        {
+            sum += nums[i];
+            auto mp = smp.find(sum - k);
+            if (mp != smp.end())
+            {
+                //map里面存的一定是在前面的元素
+                //可以尝试将map的value换为数组
+                result += mp->second;
+            }
+            smp[sum]++;
+        }
+
+        return result;
+    }
+};
+```