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feat: add python and java solutions to leetcode problem: No.0645
0645.Set Mismatch
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‎solution/0600-0699/0645.Set Mismatch/README.md

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Original file line numberDiff line numberDiff line change
@@ -27,21 +27,87 @@
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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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首先使用 1 到 n 的所有数字做异或运算，然后再与数组中的所有数字异或，得到的值就是缺失数字与重复的数字异或的结果。
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接着计算中这个值中其中一个非零的位 pos。然后 pos 位是否为 1，将 nums 数组的元素分成两部分，分别异或；接着将 `1~n` 的元素也分成两部分，分别异或。得到的两部分结果分别为 a,b，即是缺失数字与重复数字。
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最后判断数组中是否存在 a 或 b，若存在 a，说明重复数字是 a，返回 `[a,b]`，否则返回 `[b,a]`。
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<!-- tabs:start -->
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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 findErrorNums(self, nums: List[int]) -> List[int]:
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res = 0
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for num in nums:
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res ^= num
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for i in range(1, len(nums) + 1):
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res ^= i
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pos = 0
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while (res & 1) == 0:
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res >>= 1
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pos += 1
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a = b = 0
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for num in nums:
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if ((num >> pos) & 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, len(nums) + 1):
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if ((i >> pos) & 1) == 0:
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a ^= i
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else:
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b ^= i
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for num in nums:
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if num == a:
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return [a, b]
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return [b, a]
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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[] findErrorNums(int[] nums) {
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int res = 0;
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for (int num : nums) {
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res ^= num;
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}
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for (int i = 1, n = nums.length; i < n + 1; ++i) {
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res ^= i;
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}
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int pos = 0;
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while ((res & 1) == 0) {
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res >>= 1;
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++pos;
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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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if (((num >> pos) & 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, n = nums.length; i < n + 1; ++i) {
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if (((i >> pos) & 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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for (int num : nums) {
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if (num == a) {
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return new int[]{a, b};
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}
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}
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return new int[]{b, a};
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}
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}
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```
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### **...**

‎solution/0600-0699/0645.Set Mismatch/README_EN.md

Lines changed: 63 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -56,13 +56,74 @@ Given an array <code>nums</code> representing the data status of this set after
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### **Python3**
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```python
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class Solution:
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def findErrorNums(self, nums: List[int]) -> List[int]:
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res = 0
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for num in nums:
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res ^= num
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for i in range(1, len(nums) + 1):
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res ^= i
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pos = 0
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while (res & 1) == 0:
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res >>= 1
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pos += 1
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a = b = 0
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for num in nums:
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if ((num >> pos) & 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, len(nums) + 1):
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if ((i >> pos) & 1) == 0:
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a ^= i
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else:
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b ^= i
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for num in nums:
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if num == a:
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return [a, b]
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return [b, a]
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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[] findErrorNums(int[] nums) {
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int res = 0;
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for (int num : nums) {
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res ^= num;
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}
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for (int i = 1, n = nums.length; i < n + 1; ++i) {
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res ^= i;
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}
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int pos = 0;
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while ((res & 1) == 0) {
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res >>= 1;
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++pos;
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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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if (((num >> pos) & 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, n = nums.length; i < n + 1; ++i) {
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if (((i >> pos) & 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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for (int num : nums) {
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if (num == a) {
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return new int[]{a, b};
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}
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}
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return new int[]{b, a};
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}
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}
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```
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### **...**

‎solution/0600-0699/0645.Set Mismatch/Solution.java

Lines changed: 37 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,37 @@
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class Solution {
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public int[] findErrorNums(int[] nums) {
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int res = 0;
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for (int num : nums) {
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res ^= num;
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}
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for (int i = 1, n = nums.length; i < n + 1; ++i) {
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res ^= i;
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}
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int pos = 0;
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while ((res & 1) == 0) {
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res >>= 1;
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++pos;
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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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if (((num >> pos) & 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, n = nums.length; i < n + 1; ++i) {
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if (((i >> pos) & 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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for (int num : nums) {
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if (num == a) {
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return new int[]{a, b};
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}
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}
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return new int[]{b, a};
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}
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}

‎solution/0600-0699/0645.Set Mismatch/Solution.py

Lines changed: 26 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,26 @@
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class Solution:
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def findErrorNums(self, nums: List[int]) -> List[int]:
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res = 0
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for num in nums:
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res ^= num
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for i in range(1, len(nums) + 1):
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res ^= i
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pos = 0
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while (res & 1) == 0:
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res >>= 1
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pos += 1
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a = b = 0
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for num in nums:
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if ((num >> pos) & 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, len(nums) + 1):
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if ((i >> pos) & 1) == 0:
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a ^= i
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else:
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b ^= i
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for num in nums:
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if num == a:
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return [a, b]
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return [b, a]

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