add solution

Change-Id: Idbe5c6725a68b1afa0983f8a4444ea7fdd6b1d90

Author: JINGUIWANG

StringtoInteger(atoi).h

@@ -1,10 +1,10 @@
 /*
  Author:     Annie Kim, [email protected]
  Date:       May 17, 2013
- Update:     Sep 22, 2013
+ Update:     Dec 14, 2014(By [email protected])
  Problem:    String to Integer (atoi)
  Difficulty: Easy
- Source:     http://leetcode.com/onlinejudge#question_8
+ Source:     https://oj.leetcode.com/problems/string-to-integer-atoi/solution/
  Notes:
  Implement atoi to convert a string to an integer.
  Hint: Carefully consider all possible input cases. If you want a challenge, please do not 
@@ -25,12 +25,16 @@
  If no valid conversion could be performed, a zero value is returned. If the correct value is out 
  of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
 
- Solution: ...
+ Solution: 1. use long type to store the result to deal with overflow.
+           2. To deal with overflow, inspect the current number before multiplication. 
+            If the current number is greater than 214748364, we know it is going to overflow.
+            On the other hand, if the current number is equal to 214748364, 
+            we know that it will overflow only when the current digit is greater than or equal to 8..
  */
 
 class Solution {
 public:
-    int atoi(const char *str) {
+    int atoi_1(const char *str) {
         if (!str) return 0;
         while (*str == ' ') str++;
         bool positive = true;
@@ -48,4 +52,22 @@ class Solution {
         if (res < INT_MIN) return INT_MIN;
         return (int)res;
     }
+    int atoi_2(const char *str) {
+        if(str == NULL) return 0;
+        int res = 0;
+        bool sign = true;
+        while(*str == ' ') str++;
+        if(*str == '+' || *str == '-') {
+            sign = *str == '+';
+            str++;
+        }
+        while(isdigit(*str)) {
+            if(res > INT_MAX/10 || (res == INT_MAX / 10 && *str - '0' > INT_MAX % 10)){
+                return sign ? INT_MAX : INT_MIN;
+            }
+            res =  res * 10 + *str - '0';
+            str++;
+        }
+        return  sign ? res : -res;
+    }
 };