Adding CPP Implementations

src/main/cpp/leetcode/StrToInt.cpp

@@ -0,0 +1,36 @@
+/* Adding CPP implementation of the problem: https://oj.leetcode.com/problems/string-to-integer-atoi/ */
+
+/*
+*Implementing function to convert string to an integer.
+
+*If no valid conversion could be performed, a zero value is returned.
+
+*If the correct value is out of the range of representable values, the maximum integer value (2147483647) or the minimum integer value (–2147483648) is returned.
+
+*For more explanation, visit- https://www.gyanblog.com/gyan/coding-interview/leetcode-string-integer-atoi/
+*/
+
+class Solution {
+public:
+    int myAtoi(string str) {
+        int sign=1, i=0;
+        int maxdiv = INT_MAX/10;
+        while (str[i]==' ')     i++;
+        if (i<str.length() && str[i]=='+') i++;
+        else if (i<str.length() && str[i]=='-') {
+            sign=-1;
+            i++;
+        }
+        int num=0;
+        while (i<str.length()) {
+            if (str[i]<'0' || str[i]>'9')
+                return sign*num;
+            int digit = str[i]-'0';
+            if (num > maxdiv || num==maxdiv && digit>=8)
+                return sign==1 ? INT_MAX : INT_MIN;
+            num = num*10 + digit;
+            i++;
+        }
+        return sign*num;
+    }
+};

src/main/cpp/leetcode/strstr.cpp

@@ -0,0 +1,20 @@
+/* Adding CPP implementation of Problem : https://oj.leetcode.com/problems/implement-strstr/ */
+
+/* Assuming n=length of haystack, m=length of needle, then -
+O(nm) runtime, O(1) space complexity */
+
+/* BRUTE FORCE: Scan the needle with the haystack from its first position and start matching all subsequent letters one by one.
+If one of the letters does not match, we start over again with the next position in the haystack. */
+
+class Solution {
+public:
+    int strStr(string haystack, string needle) {
+        for (int i=0; ; i++) {
+            for (int j=0; ; j++) {
+                if (j == needle.length())   return i;
+                if (i+j == haystack.length())   return -1;
+                if (needle[j] != haystack[i+j])     break;
+            }
+        }      
+    }
+};