add MissingNumber

Author: TonyDoen

hi-leetcode/README.md

@@ -34,6 +34,7 @@
 | 0134 | [Gas Station](https://leetcode.com/problems/gas-station/) | [Java](./src/main/java/me/meet/leetcode/medium/GasStation.java) | Medium |
 | 0150 | [Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation/) | [Java](./src/main/java/me/meet/leetcode/medium/EvaluateReversePolishNotation.java) | Medium |
 | 0241 | [Different Ways To Add Parentheses](https://leetcode.com/problems/different-ways-to-add-parentheses/) | [Java](./src/main/java/me/meet/leetcode/medium/DifferentWaysToAddParentheses.java) | Medium |
+| 0268 | [Missing Number](https://leetcode.com/problems/missing-number/) | [Java](./src/main/java/me/meet/leetcode/medium/MissingNumber.java) | Medium |
 | 0442 | [Find All Duplicates In An Array](https://leetcode.com/problems/find-all-duplicates-in-an-array/) | [Java](./src/main/java/me/meet/leetcode/medium/FindAllDuplicatesInAnArray.java) | Medium |
 | 0445 | [Add Two Numbers II](https://leetcode.com/problems/add-two-numbers-ii/) | [Java](./src/main/java/me/meet/leetcode/medium/AddTwoNumbersII.java) | Medium |
 | 0450 | [Delete Node In BST](https://leetcode.com/problems/delete-node-in-a-bst/) | [Java](./src/main/java/me/meet/leetcode/medium/DeleteNodeInBST.java) | Medium |

hi-leetcode/src/main/java/me/meet/leetcode/medium/MissingNumber.java

@@ -0,0 +1,57 @@
+package me.meet.leetcode.medium;
+
+public final class MissingNumber {
+    private MissingNumber() {}
+
+    /**
+     * 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?
+     *
+     *
+     * 题意：
+     * 这道题给我们n个数字，是0到n之间的数但是有一个数字去掉了，让我们寻找这个数字，要求线性的时间复杂度和常数级的空间复杂度。
+     *
+     * 那么最直观的一个方法是用等差数列的求和公式求出0到n之间所有的数字之和，然后再遍历数组算出给定数字的累积和，然后做减法，差值就是丢失的那个数字
+     */
+    static int getMissingNumber(int[] arr) {
+        int sum = 0, n = arr.length;
+        for (int i = 0; i < n; i++) {
+            sum += arr[i];
+        }
+
+        return n*(n+1)/2 - sum;
+    }
+
+    /**
+     * 这题还有一种解法，使用位操作Bit Manipulation来解的，用到了异或操作的特性
+     * 那么思路是既然0到n之间少了一个数，我们将这个少了一个数的数组合0到n之间完整的数组异或一下，那么相同的数字都变为0了，剩下的就是少了的那个数字了
+     */
+    static int getMissingNumberXOR(int[] arr) {
+        int res = 0, n = arr.length;
+        for (int i = 0; i < n; i++) {
+            res ^= (i+1)^arr[i];
+        }
+        return res;
+    }
+
+    /**
+     * 1. 排序
+     * 2. 二分查找到 丢失到数字
+     */
+
+    private static void testGetMissingNumber() {
+        int[] arr = new int[]{0,1,2,3,4,6,7,8,9};
+
+        System.out.println(getMissingNumber(arr));
+        System.out.println(getMissingNumberXOR(arr));
+    }
+
+    public static void main(String[] args) {
+        testGetMissingNumber();
+    }
+}