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feat: add solutions to lcp problems: No.06,07 (#1649)
* No.06.拿硬币 * No.07.传递信息
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‎lcp/LCP 06. 拿硬币/README.md

+45-30
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@@ -35,6 +35,14 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：数学**
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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。
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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。
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时间复杂度 $O(n)$，其中 $n$ 是数组 $coins$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def minCount(self, coins: List[int]) -> int:
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return sum((coin + 1) // 2 for coin in coins)
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return sum((x + 1) >> 1 for x in coins)
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```
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### **Java**
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class Solution {
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public int minCount(int[] coins) {
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int ans = 0;
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for (int coin : coins) {
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ans += (coin + 1) / 2;
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for (int x : coins) {
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ans += (x + 1) >> 1;
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}
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return ans;
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}
@@ -70,7 +78,9 @@ class Solution {
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public:
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int minCount(vector<int>& coins) {
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int ans = 0;
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for (int coin : coins) ans += (coin + 1) / 2;
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for (int x : coins) {
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ans += (x + 1) >> 1;
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}
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return ans;
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}
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};
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### **Go**
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```go
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func minCount(coins []int) int {
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ans := 0
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for _, coin := range coins {
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ans += (coin + 1) / 2
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func minCount(coins []int) (ans int) {
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for _, x := range coins {
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ans += (x + 1) >> 1
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}
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return ans
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return
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}
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```
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### **C**
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### **TypeScript**
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```c
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int minCount(int* coins, int coinsSize) {
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int res = 0;
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for (int i = 0; i < coinsSize; i++) {
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int coin = coins[i];
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if (coin % 2 == 1) {
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res++;
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}
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res += coin / 2;
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```ts
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function minCount(coins: number[]): number {
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let ans = 0;
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for (const x of coins) {
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ans += (x + 1) >> 1;
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}
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return res;
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return ans;
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}
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```
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### **TypeScript**
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### **Rust**
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```ts
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function minCount(coins: number[]): number {
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let ans = 0;
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for (const coin of coins) {
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ans += Math.floor((coin + 1) / 2);
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```rust
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impl Solution {
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pub fn min_count(coins: Vec<i32>) -> i32 {
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coins.iter().map(|&x| (x + 1) >> 1).sum::<i32>()
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}
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}
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```
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### **C**
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```c
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int minCount(int* coins, int coinsSize) {
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int ans = 0;
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for (int i = 0; i < coinsSize; ++i) {
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ans += (coins[i] + 1) >> 1;
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}
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return ans;
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}
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* @return Integer
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*/
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function minCount($coins) {
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$cnt = 0;
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for ($i = 0; $i < count($coins); $i++) {
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$cnt += floor($coins[$i] / 2) + ($coins[$i] % 2);
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$ans = 0;
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foreach ($coins as $x) {
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$ans += $x + 1 >> 1;
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}
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return $cnt;
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return $ans;
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}
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}
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```

‎lcp/LCP 06. 拿硬币/Solution.c

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

‎lcp/LCP 06. 拿硬币/Solution.cpp

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

‎lcp/LCP 06. 拿硬币/Solution.go

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

‎lcp/LCP 06. 拿硬币/Solution.java

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

‎lcp/LCP 06. 拿硬币/Solution.php

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

‎lcp/LCP 06. 拿硬币/Solution.py

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

‎lcp/LCP 06. 拿硬币/Solution.rs

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

‎lcp/LCP 06. 拿硬币/Solution.ts

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

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