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+# [278. First Bad Version](https://leetcode.com/problems/first-bad-version/description/)
+# 思路
+思路很简单，就是二分法。    
+但是这题会超时，其实问题不是时间复杂度高(二分法的时间复杂度已经是理论最低了)，而是因为在计算`mid = (low + high) / 2`时low + high会溢出而产生不可预料的值。   
+所以，我们不应该用`mid = (high + low) / 2`来更新mid而应该`mid = low + (high - low) / 2`。   
+> **以后的二分法都应该这样更新mid以防溢出！！！**
+# C++
+```
+// Forward declaration of isBadVersion API.
+bool isBadVersion(int version);
+
+class Solution {
+public:
+    int firstBadVersion(int n) {
+        int low = 1, high = n, mid;
+        while(low <= high){
+            mid = low + (high - low) / 2; // mid = (high + low) / 2 会溢出！！！
+            if(isBadVersion(mid)) high = mid - 1;
+            else low = mid + 1;
+        }
+        return low;
+    }
+};
+```