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Commit e6527db

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feat: add solutions to lc problems: No.0268,0645 (doocs#1803)
* No.0268.Missing Number * No.0645.Set Mismatch
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‎solution/0100-0199/0136.Single Number/README.md‎

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我们对该数组所有元素进行异或运算,结果就是那个只出现一次的数字。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 是数组 `nums` 的长度。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 是数组 $nums$ 的长度。
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‎solution/0100-0199/0136.Single Number/README_EN.md‎

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## Solutions
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**Solution 1: Bitwise Operation**
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The XOR operation has the following properties:
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- Any number XOR 0 is still the original number, i.e., $x \oplus 0 = x$;
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- Any number XOR itself is 0, i.e., $x \oplus x = 0$;
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Performing XOR operation on all elements in the array will result in the number that only appears once.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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### **Python3**

‎solution/0200-0299/0268.Missing Number/README.md‎

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**方法一:位运算**
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对于数组中的每个元素,都可以与下标进行异或运算,最终的结果就是缺失的数字。
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异或运算的性质:
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- 任何数和 0ドル$ 做异或运算,结果仍然是原来的数,即 $x \oplus 0 = x$;
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- 任何数和其自身做异或运算,结果是 0ドル,ドル即 $x \oplus x = 0$;
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因此,我们可以遍历数组,将数字 $[0,..n]$ 与数组中的元素进行异或运算,最后的结果就是缺失的数字。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组长度。
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**方法二:数学**
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我们也可以用数学求解。求出 $[0,..n]$ 的和,减去数组中所有数的和,就得到了缺失的数字。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组长度。
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时间复杂度 $O(n),ドル其中 $n$ 为数组长度。空间复杂度 $O(1)$。
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn missing_number(nums: Vec<i32>) -> i32 {
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let n = nums.len() as i32;
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let mut ans = n;
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for (i, v) in nums.iter().enumerate() {
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ans ^= i as i32 ^ v;
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}
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ans
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}
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}
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```
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```rust
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impl Solution {
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pub fn missing_number(nums: Vec<i32>) -> i32 {
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let n = nums.len() as i32;
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let mut ans = n;
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for (i, &v) in nums.iter().enumerate() {
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ans += i as i32 - v;
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}
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ans
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}
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}
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```
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### **TypeScript**
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```ts
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function missingNumber(nums: number[]): number {
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const n = nums.length;
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let ans = n;
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for (let i = 0; i < n; ++i) {
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ans ^= i ^ nums[i];
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}
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return ans;
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}
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```
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```ts
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function missingNumber(nums: number[]): number {
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const n = nums.length;
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let ans = n;
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for (let i = 0; i < n; ++i) {
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ans += i - nums[i];
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}
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return ans;
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}
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```
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### **JavaScript**
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```js

‎solution/0200-0299/0268.Missing Number/README_EN.md‎

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## Solutions
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**Solution 1: Bitwise Operation**
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The XOR operation has the following properties:
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- Any number XOR 0 is still the original number, i.e., $x \oplus 0 = x$;
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- Any number XOR itself is 0, i.e., $x \oplus x = 0$;
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Therefore, we can traverse the array, perform XOR operation between each element and the numbers $[0,..n],ドル and the final result will be the missing number.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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**Solution 2: Mathematics**
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We can also solve this problem using mathematics. By calculating the sum of $[0,..n],ドル subtracting the sum of all numbers in the array, we can obtain the missing number.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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### **Python3**
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn missing_number(nums: Vec<i32>) -> i32 {
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let n = nums.len() as i32;
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let mut ans = n;
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for (i, v) in nums.iter().enumerate() {
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ans ^= i as i32 ^ v;
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}
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ans
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}
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}
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```
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```rust
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impl Solution {
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pub fn missing_number(nums: Vec<i32>) -> i32 {
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let n = nums.len() as i32;
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let mut ans = n;
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for (i, &v) in nums.iter().enumerate() {
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ans += i as i32 - v;
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}
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ans
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}
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}
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```
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### **TypeScript**
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```ts
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function missingNumber(nums: number[]): number {
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const n = nums.length;
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let ans = n;
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for (let i = 0; i < n; ++i) {
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ans ^= i ^ nums[i];
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}
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return ans;
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}
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```
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```ts
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function missingNumber(nums: number[]): number {
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const n = nums.length;
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let ans = n;
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for (let i = 0; i < n; ++i) {
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ans += i - nums[i];
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}
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return ans;
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}
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```
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### **JavaScript**
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```js
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impl Solution {
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pub fn missing_number(nums: Vec<i32>) -> i32 {
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let n = nums.len() as i32;
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let mut ans = n;
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for (i, v) in nums.iter().enumerate() {
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ans ^= i as i32 ^ v;
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}
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ans
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}
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}
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function missingNumber(nums: number[]): number {
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const n = nums.length;
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let ans = n;
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for (let i = 0; i < n; ++i) {
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ans ^= i ^ nums[i];
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}
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return ans;
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}

‎solution/0200-0299/0287.Find the Duplicate Number/README_EN.md‎

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## Solutions
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**Solution 1: Binary Search**
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We can observe that if the number of elements in $[1,..x]$ is greater than $x,ドル then the duplicate number must be in $[1,..x],ドル otherwise the duplicate number must be in $[x+1,..n]$.
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Therefore, we can use binary search to find $x,ドル and check whether the number of elements in $[1,..x]$ is greater than $x$ at each iteration. This way, we can determine which interval the duplicate number is in, and narrow down the search range until we find the duplicate number.
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The time complexity is $O(n \times \log n),ドル where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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### **Python3**

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