进一步完善[Reverse Integer]

C++/chapImplement.tex

@@ -33,15 +33,26 @@ \subsubsection{代码}
 \begin{Code}
 //LeetCode, Reverse Integer
 // 时间复杂度O(logn)，空间复杂度O(1)
+// 考虑 1.负数的情况 2. 溢出的情况(正溢出&&负溢出，比如 x = -2147483648(即-2^31) )
 class Solution {
 public:
     int reverse (int x) {
-        int r = 0;
-
-        for (; x; x /= 10)
-            r = r * 10 + x % 10;
-
-        return r;
+        long long r = 0;
+	long long t = x;
+	t = t > 0 ? t : -t;
+        for (; t; t /= 10)
+            r = r * 10 + t % 10;
+
+	bool sign = x > 0 ? false: true;
+	if (r > 2147483647 || (sign && r > 2147483648)) {
+	    return 0;
+	} else {
+	    if (sign) {
+	        return -r;
+	    } else {
+	        return r;
+	    }
+	}
     }
 };
 \end{Code}