From f1ae30a3f0de5e1a9de1eaa9f4c1f30be555f791 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Fri, 28 Sep 2018 14:59:29 +0800
Subject: [PATCH] Create 69. Sqrt(x).md

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+# [69. Sqrt(x)](https://leetcode.com/problems/sqrtx/description/)
+# 思路
+求int型数的平方根并向下取整。   
+考虑采用二分法。    
+注意：为了防止溢出，判断mid * mid与 x 的值时用除法，即判断 mid 与 x / mid的大小关系。   
+# C++
+```
+class Solution {
+public:
+    int mySqrt(int x) {
+        if(x < 2) return x;
+        int low = 1, high = x / 2, mid;
+        while(low <= high){
+            mid = (low + high) / 2;
+            if(mid == x / mid) return mid;
+            else if(mid < x / mid) low = mid + 1;
+            else high = mid - 1;
+        }
+        return high;  
+    }
+};
+```