add sqrt

Author: tg123

sqrtx/README.md

@@ -1,4 +1,69 @@
+## [Newton's Method](http://en.wikipedia.org/wiki/Newton%27s_method)
+
+This is copied from my friend Newton.
+
+Then I wrote this code:
+
+```
+public int sqrt(int x) {
+    if(x < 0) return -1;
+    if(x == 0) return 0;
+
+    // x/t + t = 2t
+    double t = (double)x;
+    double tt = 0;
+    while( (int)tt - (int)t != 0 ){
+        tt = t;
+        t = (x/t + t) /2;
+        
+    }
+    
+    return (int)tt;
+}
+```
+
+## Programming Way
+
+Well, I replaced the mathematical way finally, maybe, I think I should do this in programming way.
+
+### Brute Force
+
+Search through `1..x`, find the `i`, such that `i^2 == x` or  `i^2 < x < (i + 1)^2`.
+
+
+### Binary Search
+
+As we can brute force, and `1..x` is obviously ordered, we can do it by applying binary search pattern.
+
+
+```
+treat 1..x as the content of an array[0..x-1]
+
+
+while s < e
+  
+  m = (s + e) / 2
+  
+  if (m + 1)^2 == x
+    return m + 1
+  
+  if (m + 1)^2 < x < (m + 1 + 1)^2
+    return m + 1
+  
+  if x < (m + 1)^2
+    e = m
+  else
+    s = m + 1
+```
+
+#### Overflow
+
+The testcases contain some large numbers like `Integer.MAX_VALUE`, code will fail due to overflow.
+
+To avoid this, change any `+` and `*` to `-` and `/`
+  
+  * `m = (s + e) / 2` -> `m = (e - s) / 2 + s`
+  * `(m + 1)^2 == x`  -> `x / (m + 1) == (m + 1)`
+
 
-## TODO 
-  * write down thinking