feat: add solutions to lcp problem: No.06

Author: yanglbme

lcp/LCP 06. 拿硬币/README.md

@@ -42,15 +42,50 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minCount(self, coins: List[int]) -> int:
+        return sum((coin + 1) // 2 for coin in coins)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int minCount(int[] coins) {
+        int ans = 0;
+        for (int coin : coins) {
+            ans += (coin + 1) / 2;
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minCount(vector<int>& coins) {
+        int ans = 0;
+        for (int coin : coins) ans += (coin + 1) / 2;
+        return ans;
+    }
+};
+```
+
+### **Go**
 
+```go
+func minCount(coins []int) int {
+	ans := 0
+	for _, coin := range coins {
+		ans += (coin + 1) / 2
+	}
+	return ans
+}
 ```
 
 ### **...**

lcp/LCP 06. 拿硬币/Solution.cpp

@@ -0,0 +1,8 @@
+class Solution {
+public:
+    int minCount(vector<int>& coins) {
+        int ans = 0;
+        for (int coin : coins) ans += (coin + 1) / 2;
+        return ans;
+    }
+};

lcp/LCP 06. 拿硬币/Solution.go

@@ -0,0 +1,7 @@
+func minCount(coins []int) int {
+	ans := 0
+	for _, coin := range coins {
+		ans += (coin + 1) / 2
+	}
+	return ans
+}

lcp/LCP 06. 拿硬币/Solution.java

@@ -0,0 +1,9 @@
+class Solution {
+    public int minCount(int[] coins) {
+        int ans = 0;
+        for (int coin : coins) {
+            ans += (coin + 1) / 2;
+        }
+        return ans;
+    }
+}

lcp/LCP 06. 拿硬币/Solution.py

@@ -0,0 +1,3 @@
+class Solution:
+    def minCount(self, coins: List[int]) -> int:
+        return sum((coin + 1) // 2 for coin in coins)

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

@@ -85,7 +85,6 @@ class Solution {
         int eor = 0;
         for (int i = 1; i <= nums.length; ++i) {
             eor ^= (i ^ nums[i - 1]);
-<<<<<<< Updated upstream
         }
         int diff = eor & (~eor + 1);
         int a = 0;
@@ -158,27 +157,14 @@ public:
             if ((nums[i - 1] & diff) == 0) {
                 a ^= nums[i - 1];
             }
-=======
-        }
-        int diff = eor & (~eor + 1);
-        int a = 0;
-        for (int i = 1; i <= nums.length; ++i) {
-            if ((nums[i - 1] & diff) == 0) {
-                a ^= nums[i - 1];
-            }
->>>>>>> Stashed changes
             if ((i & diff) == 0) {
                 a ^= i;
             }
         }
         int b = eor ^ a;
         for (int num : nums) {
             if (a == num) {
-<<<<<<< Updated upstream
                 return {a, b};
-=======
-                return new int[]{a, b};
->>>>>>> Stashed changes
             }
         }
         return {b, a};

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

@@ -61,7 +61,6 @@ class Solution {
         int eor = 0;
         for (int i = 1; i <= nums.length; ++i) {
             eor ^= (i ^ nums[i - 1]);
-<<<<<<< Updated upstream
         }
         int diff = eor & (~eor + 1);
         int a = 0;
@@ -134,27 +133,14 @@ public:
             if ((nums[i - 1] & diff) == 0) {
                 a ^= nums[i - 1];
             }
-=======
-        }
-        int diff = eor & (~eor + 1);
-        int a = 0;
-        for (int i = 1; i <= nums.length; ++i) {
-            if ((nums[i - 1] & diff) == 0) {
-                a ^= nums[i - 1];
-            }
->>>>>>> Stashed changes
             if ((i & diff) == 0) {
                 a ^= i;
             }
         }
         int b = eor ^ a;
         for (int num : nums) {
             if (a == num) {
-<<<<<<< Updated upstream
                 return {a, b};
-=======
-                return new int[]{a, b};
->>>>>>> Stashed changes
             }
         }
         return {b, a};
@@ -164,6 +150,8 @@ public:
 
 ### **Go**
 
+把每个数都放到它应该在的位置，最后出现“异常”的就是重复的数和丢失的数。
+
 ```go
 func findErrorNums(nums []int) []int {
 	n := len(nums)
@@ -181,6 +169,34 @@ func findErrorNums(nums []int) []int {
 }
 ```
 
+也可以使用位运算。
+
+```go
+func findErrorNums(nums []int) []int {
+	eor, n := 0, len(nums)
+	for i := 1; i <= n; i++ {
+		eor ^= (i ^ nums[i-1])
+	}
+	diff := eor & (-eor)
+	a := 0
+	for i := 1; i <= n; i++ {
+		if (nums[i-1] & diff) == 0 {
+			a ^= nums[i-1]
+		}
+		if (i & diff) == 0 {
+			a ^= i
+		}
+	}
+	b := eor ^ a
+	for _, num := range nums {
+		if a == num {
+			return []int{a, b}
+		}
+	}
+	return []int{b, a}
+}
+```
+
 ### **C++**
 
 ```cpp