feat: add solutions to lcp problems: No.06,07 (#1649)

* No.06.拿硬币
* No.07.传递信息

Author: yanglbme

lcp/LCP 06. 拿硬币/README.md

@@ -35,6 +35,14 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：数学**
+
+我们可以发现，每堆力扣币拿完的最少次数，等于该堆力扣币数量除以 $2$ 向上取整的结果之和。
+
+因此，我们只需要遍历每堆力扣币 $x_i$，计算每堆力扣币拿完的最少次数 $\left \lceil x_i/2 \right \rceil$，然后累加即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是数组 $coins$ 的长度。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -44,7 +52,7 @@
 ```python
 class Solution:
     def minCount(self, coins: List[int]) -> int:
-        return sum((coin + 1) // 2 for coin in coins)
+        return sum((x + 1) >> 1 for x in coins)
 ```
 
 ### **Java**
@@ -55,8 +63,8 @@ class Solution:
 class Solution {
     public int minCount(int[] coins) {
         int ans = 0;
-        for (int coin : coins) {
-            ans += (coin + 1) / 2;
+        for (int x : coins) {
+            ans += (x + 1) >> 1;
         }
         return ans;
     }
@@ -70,7 +78,9 @@ class Solution {
 public:
     int minCount(vector<int>& coins) {
         int ans = 0;
-        for (int coin : coins) ans += (coin + 1) / 2;
+        for (int x : coins) {
+            ans += (x + 1) >> 1;
+        }
         return ans;
     }
 };
@@ -79,38 +89,43 @@ public:
 ### **Go**
 
 ```go
-func minCount(coins []int) int {
-	ans := 0
-	for _, coin := range coins {
-		ans += (coin + 1) / 2
+func minCount(coins []int) (ans int) {
+	for _, x := range coins {
+		ans += (x + 1) >> 1
 	}
-	return ans
+	return
 }
 ```
 
-### **C**
+### **TypeScript**
 
-```c
-int minCount(int* coins, int coinsSize) {
-    int res = 0;
-    for (int i = 0; i < coinsSize; i++) {
-        int coin = coins[i];
-        if (coin % 2 == 1) {
-            res++;
-        }
-        res += coin / 2;
+```ts
+function minCount(coins: number[]): number {
+    let ans = 0;
+    for (const x of coins) {
+        ans += (x + 1) >> 1;
     }
-    return res;
+    return ans;
 }
 ```
 
-### **TypeScript**
+### **Rust**
 
-```ts
-function minCount(coins: number[]): number {
-    let ans = 0;
-    for (const coin of coins) {
-        ans += Math.floor((coin + 1) / 2);
+```rust
+impl Solution {
+    pub fn min_count(coins: Vec<i32>) -> i32 {
+        coins.iter().map(|&x| (x + 1) >> 1).sum::<i32>()
+    }
+}
+```
+
+### **C**
+
+```c
+int minCount(int* coins, int coinsSize) {
+    int ans = 0;
+    for (int i = 0; i < coinsSize; ++i) {
+        ans += (coins[i] + 1) >> 1;
     }
     return ans;
 }
@@ -125,11 +140,11 @@ class Solution {
      * @return Integer
      */
     function minCount($coins) {
-        $cnt = 0;
-        for ($i = 0; $i < count($coins); $i++) {
-            $cnt += floor($coins[$i] / 2) + ($coins[$i] % 2);
+        $ans = 0;
+        foreach ($coins as $x) {
+            $ans += $x + 1 >> 1;
         }
-        return $cnt;
+        return $ans;
     }
 }
 ```

lcp/LCP 06. 拿硬币/Solution.c

@@ -1,11 +1,7 @@
-int minCount(int* coins, int coinsSize) {
-    int res = 0;
-    for (int i = 0; i < coinsSize; i++) {
-        int coin = coins[i];
-        if (coin % 2 == 1) {
-            res++;
-        }
-        res += coin / 2;
-    }
-    return res;
-}
+int minCount(int* coins, int coinsSize) {
+    int ans = 0;
+    for (int i = 0; i < coinsSize; ++i) {
+        ans += (coins[i] + 1) >> 1;
+    }
+    return ans;
+}

lcp/LCP 06. 拿硬币/Solution.cpp

@@ -1,8 +1,10 @@
-class Solution {
-public:
-    int minCount(vector<int>& coins) {
-        int ans = 0;
-        for (int coin : coins) ans += (coin + 1) / 2;
-        return ans;
-    }
+class Solution {
+public:
+    int minCount(vector<int>& coins) {
+        int ans = 0;
+        for (int x : coins) {
+            ans += (x + 1) >> 1;
+        }
+        return ans;
+    }
 };

lcp/LCP 06. 拿硬币/Solution.go

@@ -1,7 +1,6 @@
-func minCount(coins []int) int {
-	ans := 0
-	for _, coin := range coins {
-		ans += (coin + 1) / 2
+func minCount(coins []int) (ans int) {
+	for _, x := range coins {
+		ans += (x + 1) >> 1
 	}
-	return ans
+	return
 }

lcp/LCP 06. 拿硬币/Solution.java

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

lcp/LCP 06. 拿硬币/Solution.php

@@ -4,10 +4,10 @@ class Solution {
      * @return Integer
      */
     function minCount($coins) {
-        $cnt = 0;
-        for ($i = 0; $i < count($coins); $i++) {
-            $cnt += floor($coins[$i] / 2) + ($coins[$i] % 2);
+        $ans = 0;
+        foreach ($coins as $x) {
+            $ans += $x + 1 >> 1;
         }
-        return $cnt;
+        return $ans;
     }
 }

lcp/LCP 06. 拿硬币/Solution.py

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

lcp/LCP 06. 拿硬币/Solution.rs

@@ -0,0 +1,5 @@
+impl Solution {
+    pub fn min_count(coins: Vec<i32>) -> i32 {
+        coins.iter().map(|&x| (x + 1) >> 1).sum::<i32>()
+    }
+}

lcp/LCP 06. 拿硬币/Solution.ts

@@ -1,7 +1,7 @@
 function minCount(coins: number[]): number {
     let ans = 0;
-    for (const coin of coins) {
-        ans += Math.floor((coin + 1) / 2);
+    for (const x of coins) {
+        ans += (x + 1) >> 1;
     }
     return ans;
 }