Merge branch 'main' of https://github.91chifun.workers.dev//https://github.com/doocs/leetcode into main

addd new cpp solution for leetcode 322 [Coin Change]

Author: junfeng.fj

lcof/面试题56 - I. 数组中数字出现的次数/README.md

@@ -26,7 +26,11 @@
 
 异或运算求解。
 
-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。
+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**，即 `eor = a ^ b`
+
+找出这个结果 eor 中最后一个二进制位为 1 而其余位为 0 的数，即 `eor & (~eor + 1)`，之后遍历数组所有元素，二进制位为 0 的元素异或到 a。
+
+遍历结束后 `b = eor ^ a`，返回结果即可。
 
 <!-- tabs:start -->
 
@@ -35,21 +39,16 @@
 ```python
 class Solution:
     def singleNumbers(self, nums: List[int]) -> List[int]:
-        xor_res = 0
+        eor = 0
         for num in nums:
-            xor_res ^= num
-        pos = 0
-        while (xor_res & 1) == 0:
-            pos += 1
-            xor_res >>= 1
-
-        a = b = 0
+            eor ^= num
+        # 找出最右边的 1
+        diff = eor & (~eor + 1)
+        a = 0
         for num in nums:
-            t = num >> pos
-            if (t & 1) == 0:
+            if (num & diff) == 0:
                 a ^= num
-            else:
-                b ^= num
+        b = eor ^ a
         return [a, b]
 ```
 
@@ -58,25 +57,20 @@ class Solution:
 ```java
 class Solution {
     public int[] singleNumbers(int[] nums) {
-        int xor = 0;
+        int eor = 0;
         for (int num : nums) {
-            xor ^= num;
+            eor ^= num;
         }
-        int pos = 0;
-        while ((xor & 1) == 0) {
-            ++pos;
-            xor >>= 1;
-        }
-        int a = 0, b = 0;
+        // # 找出最右边的 1
+        int diff = eor & (~eor + 1);
+        int a = 0;
         for (int num : nums) {
-            int t = num >> pos;
-            if ((t & 1) == 0) {
+            if ((num & diff) == 0) {
                 a ^= num;
-            } else {
-                b ^= num;
             }
         }
-        return new int[] {a, b};
+        int b = eor ^ a;
+        return new int[]{a, b};
     }
 }
 ```
@@ -89,24 +83,19 @@ class Solution {
  * @return {number[]}
  */
 var singleNumbers = function (nums) {
-  let xor = 0;
-  let bit = 1;
-  let res = [0, 0];
+  let eor = 0;
   for (let num of nums) {
-    xor ^= num;
-  }
-  while ((xor & 1) === 0) {
-    xor >>= 1;
-    bit <<= 1;
+    eor ^= num;
   }
+  const diff = eor & (~eor + 1);
+  let a = 0;
   for (let num of nums) {
-    if ((num & bit) === 0) {
-      res[0] ^= num;
-    } else {
-      res[1] ^= num;
+    if ((num & diff) == 0) {
+      a ^= num;
     }
   }
-  return res;
+  let b = eor ^ a;
+  return [a, b];
 };
 ```
 

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

@@ -1,23 +1,18 @@
 class Solution {
     public int[] singleNumbers(int[] nums) {
-        int xor = 0;
+        int eor = 0;
         for (int num : nums) {
-            xor ^= num;
+            eor ^= num;
         }
-        int pos = 0;
-        while ((xor & 1) == 0) {
-            ++pos;
-            xor >>= 1;
-        }
-        int a = 0, b = 0;
+        // # 找出最右边的 1
+        int diff = eor & (~eor + 1);
+        int a = 0;
         for (int num : nums) {
-            int t = num >> pos;
-            if ((t & 1) == 0) {
+            if ((num & diff) == 0) {
                 a ^= num;
-            } else {
-                b ^= num;
             }
         }
-        return new int[] { a, b };
+        int b = eor ^ a;
+        return new int[]{a, b};
     }
 }

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

@@ -3,22 +3,17 @@
  * @return {number[]}
  */
 var singleNumbers = function (nums) {
-  let xor = 0;
-  let bit = 1;
-  let res = [0, 0];
+  let eor = 0;
   for (let num of nums) {
-    xor ^= num;
-  }
-  while ((xor & 1) === 0) {
-    xor >>= 1;
-    bit <<= 1;
+    eor ^= num;
   }
+  const diff = eor & (~eor + 1);
+  let a = 0;
   for (let num of nums) {
-    if ((num & bit) === 0) {
-      res[0] ^= num;
-    } else {
-      res[1] ^= num;
+    if ((num & diff) == 0) {
+      a ^= num;
     }
   }
-  return res;
+  let b = eor ^ a;
+  return [a, b];
 };

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

@@ -1,18 +1,13 @@
 class Solution:
     def singleNumbers(self, nums: List[int]) -> List[int]:
-        xor_res = 0
+        eor = 0
         for num in nums:
-            xor_res ^= num
-        pos = 0
-        while (xor_res & 1) == 0:
-            pos += 1
-            xor_res >>= 1
-        
-        a = b = 0
+            eor ^= num
+        # 找出最右边的 1
+        diff = eor & (~eor + 1)
+        a = 0
         for num in nums:
-            t = num >> pos
-            if (t & 1) == 0:
+            if (num & diff) == 0:
                 a ^= num
-            else:
-                b ^= num
+        b = eor ^ a
         return [a, b]

solution/0200-0299/0260.Single Number III/README.md

@@ -32,17 +32,16 @@
 ```python
 class Solution:
     def singleNumber(self, nums: List[int]) -> List[int]:
-        xor = 0
+        eor = 0
         for num in nums:
-            xor ^= num
-        # x & (-x) 是保留位中最右边 1 ，且将其余的 1 设位 0 的方法
-        diff = xor & (-xor)
-        a = b = 0
+            eor ^= num
+        # 提取最右边的 1
+        diff = eor & (~eor + 1)
+        a = 0
         for num in nums:
             if (num & diff) == 0:
                 a ^= num
-            else:
-                b ^= num
+        b = eor ^ a
         return [a, b]
 ```
 
@@ -53,16 +52,19 @@ class Solution:
 ```java
 class Solution {
     public int[] singleNumber(int[] nums) {
-        int xor = 0;
+        int eor = 0;
         for (int num : nums) {
-            xor ^= num;
+            eor ^= num;
         }
-        int diff = xor & (-xor);
-        int a = 0, b = 0;
+        // 提取最右的 1
+        int diff = eor & (~eor + 1);
+        int a = 0;
         for (int num : nums) {
-            if ((num & diff) == 0) a ^= num;
-            else b ^= num;
+            if ((num & diff) == 0) {
+                a ^= num;
+            }
         }
+        int b = eor ^ a;
         return new int[]{a, b};
     }
 }

solution/0200-0299/0260.Single Number III/README_EN.md

@@ -30,35 +30,35 @@
 ```python
 class Solution:
     def singleNumber(self, nums: List[int]) -> List[int]:
-        xor = 0
+        eor = 0
         for num in nums:
-            xor ^= num
-        diff = xor & (-xor)
-        a = b = 0
+            eor ^= num
+        diff = eor & (~eor + 1)
+        a = 0
         for num in nums:
             if (num & diff) == 0:
                 a ^= num
-            else:
-                b ^= num
+        b = eor ^ a
         return [a, b]
-
 ```
 
 ### **Java**
 
 ```java
 class Solution {
     public int[] singleNumber(int[] nums) {
-        int xor = 0;
+        int eor = 0;
         for (int num : nums) {
-            xor ^= num;
+            eor ^= num;
         }
-        int diff = xor & (-xor);
-        int a = 0, b = 0;
+        int diff = eor & (~eor + 1);
+        int a = 0;
         for (int num : nums) {
-            if ((num & diff) == 0) a ^= num;
-            else b ^= num;
+            if ((num & diff) == 0) {
+                a ^= num;
+            }
         }
+        int b = eor ^ a;
         return new int[]{a, b};
     }
 }

solution/0200-0299/0260.Single Number III/Solution.java

@@ -1,15 +1,18 @@
 class Solution {
     public int[] singleNumber(int[] nums) {
-        int xor = 0;
+        int eor = 0;
         for (int num : nums) {
-            xor ^= num;
+            eor ^= num;
         }
-        int diff = xor & (-xor);
-        int a = 0, b = 0;
+        // 提取最右的 1
+        int diff = eor & (~eor + 1);
+        int a = 0;
         for (int num : nums) {
-            if ((num & diff) == 0) a ^= num;
-            else b ^= num;
+            if ((num & diff) == 0) {
+                a ^= num;
+            }
         }
+        int b = eor ^ a;
         return new int[]{a, b};
     }
 }

solution/0200-0299/0260.Single Number III/Solution.py

@@ -1,13 +1,13 @@
 class Solution:
     def singleNumber(self, nums: List[int]) -> List[int]:
-        xor = 0
+        eor = 0
         for num in nums:
-            xor ^= num
-        diff = xor & (-xor)
-        a = b = 0
+            eor ^= num
+        # 提取最右边的 1
+        diff = eor & (~eor + 1)
+        a = 0
         for num in nums:
             if (num & diff) == 0:
                 a ^= num
-            else:
-                b ^= num
+        b = eor ^ a
         return [a, b]