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Commit 2cc8917

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feat: add python and java solutions to lcci question: 17.19.Missing Two
See https://lc.netlify.app/#/lcci/17.19.Missing%20Two/README 
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‎README.md‎

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<p align="center">
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<a href="https://github.com/doocs/leetcode"><img src="https://badgen.net/badge/langs/Java,C++,Python,JavaScript,Go,.../green?list=1" alt="languages"></a>
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<a href="https://github.com/doocs/leetcode/stargazers"><img src="https://badgen.net/github/stars/doocs/leetcode" alt="stars"></a>
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<a href="https://github.com/doocs/leetcode/issues"><img src="https://badgen.net/github/open-issues/doocs/leetcode" alt="issues"></a>
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<a href="https://github.com/doocs/leetcode/network/members"><img src="https://img.shields.io/github/forks/doocs/leetcode.svg" alt="forks"></a>
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<a href="https://github.com/doocs/leetcode/network/members"><img src="https://badgen.net/github/forks/doocs/leetcode" alt="forks"></a>
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<a href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img src="https://badgen.net/github/license/doocs/leetcode?color=green" alt="LICENSE"></a><br>
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<a href="http://makeapullrequest.com"><img src="https://badgen.net/badge/PRs/welcome/cyan" alt="PRs Welcome"></a>
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<a href="https://doocs.github.io/#/?id=how-to-join"><img src="https://badgen.net/badge/organization/join%20us/cyan" alt="open-source-organization"></a>
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[English Version](/README_EN.md)
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## 站点
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- Netlify: https://lc.netlify.com
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- Netlify: https://lc.netlify.app
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- GitHub Pages: https://doocs.github.io/leetcode
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- Coding Pages: https://d3jc40.coding-pages.com
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‎README_EN.md‎

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<p align="center">
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<a href="https://github.com/doocs/leetcode"><img src="https://badgen.net/badge/langs/Java,C++,Python,JavaScript,Go,.../green?list=1" alt="languages"></a>
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<a href="https://github.com/doocs/leetcode/stargazers"><img src="https://badgen.net/github/stars/doocs/leetcode" alt="stars"></a>
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<a href="https://github.com/doocs/leetcode/issues"><img src="https://badgen.net/github/open-issues/doocs/leetcode" alt="issues"></a>
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<a href="https://github.com/doocs/leetcode/network/members"><img src="https://img.shields.io/github/forks/doocs/leetcode.svg" alt="forks"></a>
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<a href="https://github.com/doocs/leetcode/network/members"><img src="https://badgen.net/github/forks/doocs/leetcode" alt="forks"></a>
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<a href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img src="https://badgen.net/github/license/doocs/leetcode?color=green" alt="LICENSE"></a><br>
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<a href="http://makeapullrequest.com"><img src="https://badgen.net/badge/PRs/welcome/cyan" alt="PRs Welcome"></a>
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<a href="https://doocs.github.io/#/?id=how-to-join"><img src="https://badgen.net/badge/organization/join%20us/cyan" alt="open-source-organization"></a>
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[中文文档](/README.md)
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## Sites
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- Netlify: https://lc.netlify.com
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- Netlify: https://lc.netlify.app
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- GitHub Pages: https://doocs.github.io/leetcode
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- Coding Pages: https://d3jc40.coding-pages.com
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‎lcci/17.19.Missing Two/README.md‎

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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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异或运算。与[面试题56 - I. 数组中数字出现的次数](/lcof/面试题56%20-%20I.%20数组中数字出现的次数/README.md) 类似。
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### Python3
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def missingTwo(self, nums: List[int]) -> List[int]:
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res, n = 0, len(nums)
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for i in range(n):
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res ^= nums[i]
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res ^= (i + 1)
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res ^= (n + 1)
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res ^= (n + 2)
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pos = 0
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while (res & 1) == 0:
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pos += 1
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res >>= 1
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a = b = 0
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for num in nums:
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t = num >> pos
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if (t & 1) == 0:
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a ^= num
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else:
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b ^= num
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for i in range(1, n + 3):
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t = i >> pos
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if (t & 1) == 0:
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a ^= i
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else:
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b ^= i
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return [a, b]
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```
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### Java
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] missingTwo(int[] nums) {
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int res = 0, n = nums.length;
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for (int i = 0; i < n; ++i) {
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res ^= nums[i];
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res ^= (i + 1);
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}
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res ^= (n + 1);
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res ^= (n + 2);
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int pos = 0;
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while ((res & 1) == 0) {
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pos += 1;
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res >>= 1;
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}
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int a = 0, b = 0;
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for (int num : nums) {
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int t = num >> pos;
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if ((t & 1) == 0) {
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a ^= num;
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} else {
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b ^= num;
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}
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}
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for (int i = 1; i <= n + 2; ++i) {
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int t = i >> pos;
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if ((t & 1) == 0) {
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a ^= i;
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} else {
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b ^= i;
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}
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}
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return new int[]{a, b};
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}
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}
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```
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### ...

‎lcci/17.19.Missing Two/README_EN.md‎

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### Python3
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```python
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class Solution:
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def missingTwo(self, nums: List[int]) -> List[int]:
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res, n = 0, len(nums)
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for i in range(n):
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res ^= nums[i]
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res ^= (i + 1)
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res ^= (n + 1)
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res ^= (n + 2)
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pos = 0
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while (res & 1) == 0:
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pos += 1
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res >>= 1
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a = b = 0
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for num in nums:
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t = num >> pos
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if (t & 1) == 0:
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a ^= num
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else:
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b ^= num
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for i in range(1, n + 3):
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t = i >> pos
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if (t & 1) == 0:
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a ^= i
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else:
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b ^= i
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return [a, b]
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```
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### Java
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```java
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class Solution {
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public int[] missingTwo(int[] nums) {
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int res = 0, n = nums.length;
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for (int i = 0; i < n; ++i) {
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res ^= nums[i];
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res ^= (i + 1);
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}
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res ^= (n + 1);
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res ^= (n + 2);
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int pos = 0;
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while ((res & 1) == 0) {
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pos += 1;
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res >>= 1;
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}
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int a = 0, b = 0;
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for (int num : nums) {
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int t = num >> pos;
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if ((t & 1) == 0) {
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a ^= num;
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} else {
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b ^= num;
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}
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}
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for (int i = 1; i <= n + 2; ++i) {
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int t = i >> pos;
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if ((t & 1) == 0) {
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a ^= i;
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} else {
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b ^= i;
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}
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}
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return new int[]{a, b};
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}
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}
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```
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### ...

‎lcci/17.19.Missing Two/Solution.java‎

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class Solution {
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public int[] missingTwo(int[] nums) {
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int res = 0, n = nums.length;
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for (int i = 0; i < n; ++i) {
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res ^= nums[i];
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res ^= (i + 1);
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}
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res ^= (n + 1);
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res ^= (n + 2);
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int pos = 0;
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while ((res & 1) == 0) {
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pos += 1;
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res >>= 1;
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}
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int a = 0, b = 0;
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for (int num : nums) {
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int t = num >> pos;
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if ((t & 1) == 0) {
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a ^= num;
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} else {
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b ^= num;
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}
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}
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for (int i = 1; i <= n + 2; ++i) {
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int t = i >> pos;
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if ((t & 1) == 0) {
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a ^= i;
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} else {
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b ^= i;
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}
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}
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return new int[] { a, b };
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}
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}

‎lcci/17.19.Missing Two/Solution.py‎

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class Solution:
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def missingTwo(self, nums: List[int]) -> List[int]:
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res, n = 0, len(nums)
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for i in range(n):
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res ^= nums[i]
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res ^= (i + 1)
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res ^= (n + 1)
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res ^= (n + 2)
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pos = 0
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while (res & 1) == 0:
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pos += 1
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res >>= 1
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a = b = 0
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for num in nums:
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t = num >> pos
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if (t & 1) == 0:
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a ^= num
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else:
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b ^= num
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for i in range(1, n + 3):
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t = i >> pos
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if (t & 1) == 0:
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a ^= i
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else:
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b ^= i
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return [a, b]

‎lcof/面试题56 - I. 数组中数字出现的次数/README.md‎

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- `2 <= nums <= 10000`
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## 解法
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异或运算求解。
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首先明确,两个相同的数异或之后的结果为 0。对该数组所有元素进行异或运算,结果就是**两个只出现一次的数字异或的结果**。找出这个结果中某个二进制位为 1 的位置,之后对数组所有元素进行分类,二进制位为 0 的异或到 a,二进制位为 1 的异或到 b,结果就是 a,b。
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### Python3
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```python
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class Solution:

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