add solution of problem 268: Missing number

Author: ppulse

MissingNumber268/README.md

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+Given an array containing *n* distinct numbers taken from `0, 1, 2, ..., n`, find the one that is missing from the array.
+
+For example,
+Given *nums* = `[0, 1, 3]` return `2`.
+
+#####Note:
+Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
+
+
+#####思路
+0、1、2....n之和expectedSum为(n+1)*n/2
+
+拿掉一个数k后的和sum为(n+1)*n/2-k
+  
+因此k = expectedSum - sum

MissingNumber268/Solution.java

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+public class Solution {
+    public int missingNumber(int[] nums) {
+        if (nums == null || nums.length == 0)
+            return 0;
+        
+        int expectedSum = (nums.length + 1) * nums.length / 2;
+        int sum = 0;
+        for (int i = 0; i < nums.length; i++)
+            sum += nums[i];
+        
+        return expectedSum - sum;
+    }
+}