feat: add python and java solutions to leetcode problem: No.0645

0645.Set Mismatch

Author: yanglbme

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

@@ -27,21 +27,87 @@
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
 
+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。
+
+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。
+
+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]`。
 
 <!-- tabs:start -->
 
 ### **Python3**
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def findErrorNums(self, nums: List[int]) -> List[int]:
+        res = 0
+        for num in nums:
+            res ^= num
+        for i in range(1, len(nums) + 1):
+            res ^= i
+        pos = 0
+        while (res & 1) == 0:
+            res >>= 1
+            pos += 1
+        a = b = 0
+        for num in nums:
+            if ((num >> pos) & 1) == 0:
+                a ^= num
+            else:
+                b ^= num
+        for i in range(1, len(nums) + 1):
+            if ((i >> pos) & 1) == 0:
+                a ^= i
+            else:
+                b ^= i
+        for num in nums:
+            if num == a:
+                return [a, b]
+        return [b, a]
 ```
 
 ### **Java**
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] findErrorNums(int[] nums) {
+        int res = 0;
+        for (int num : nums) {
+            res ^= num;
+        }
+        for (int i = 1, n = nums.length; i < n + 1; ++i) {
+            res ^= i;
+        }
+        int pos = 0;
+        while ((res & 1) == 0) {
+            res >>= 1;
+            ++pos;
+        }
+        int a = 0, b = 0;
+        for (int num : nums) {
+            if (((num >> pos) & 1) == 0) {
+                a ^= num;
+            } else {
+                b ^= num;
+            }
+        }
+        for (int i = 1, n = nums.length; i < n + 1; ++i) {
+            if (((i >> pos) & 1) == 0) {
+                a ^= i;
+            } else {
+                b ^= i;
+            }
+        }
+        for (int num : nums) {
+            if (num == a) {
+                return new int[]{a, b};
+            }
+        }
+        return new int[]{b, a};
+    }
+}
 ```
 
 ### **...**

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

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

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

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

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

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