feat: add cpp solution to lcof problem: No.67

lcof/面试题67. 把字符串转换成整数/README.md

@@ -143,6 +143,46 @@ class Solution {
 }
 ```
 
+#### C++
+
+```cpp
+class Solution {
+public:
+    int strToInt(string str) {
+        int res = 0, bndry = INT_MAX / 10;
+        int i = 0, sign = 1, length = str.size();
+        if (length == 0) {
+            return 0;
+        }
+        // 删除首部空格
+        while (str[i] == ' ') {
+            if (++i == length) {
+                return 0;
+            }
+        }
+        // 若有负号则标识符号位
+        if (str[i] == '-') {
+            sign = -1;
+        }
+        if (str[i] == '-' || str[i] == '+') {
+            i++;
+        }
+        for (int j = i; j < length; j++) {
+            if (str[j] < '0' || str[j] > '9') {
+                break;
+            }
+            // res>214748364越界；res=214748364且str[j] > '7'越界
+            if (res > bndry || res == bndry && str[j] > '7') {
+                return sign == 1 ? INT_MAX : INT_MIN;
+            }
+            // 从左向右遍历数字并更新结果
+            res = res * 10 + (str[j] - '0');
+        }
+        return sign * res;
+    }
+};
+```
+
 #### Go
 
 ```go

lcof/面试题67. 把字符串转换成整数/Solution.cpp

@@ -0,0 +1,35 @@
+class Solution {
+public:
+    int strToInt(string str) {
+        int res = 0, bndry = INT_MAX / 10;
+        int i = 0, sign = 1, length = str.size();
+        if (length == 0) {
+            return 0;
+        }
+        // 删除首部空格
+        while (str[i] == ' ') {
+            if (++i == length) {
+                return 0;
+            }
+        }
+        // 若有负号则标识符号位
+        if (str[i] == '-') {
+            sign = -1;
+        }
+        if (str[i] == '-' || str[i] == '+') {
+            i++;
+        }
+        for (int j = i; j < length; j++) {
+            if (str[j] < '0' || str[j] > '9') {
+                break;
+            }
+            // res>214748364越界；res=214748364且str[j] > '7'越界
+            if (res > bndry || res == bndry && str[j] > '7') {
+                return sign == 1 ? INT_MAX : INT_MIN;
+            }
+            // 从左向右遍历数字并更新结果
+            res = res * 10 + (str[j] - '0');
+        }
+        return sign * res;
+    }
+};