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feat: add solutions to lcp problem: No.72
LCP 72.补给马车
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‎lcp/LCP 72. 补给马车/README.md

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远征队即将开启未知的冒险之旅，不过在此之前，将对补给车队进行最后的检查。`supplies[i]` 表示编号为 `i` 的补给马车装载的物资数量。
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考虑到车队过长容易被野兽偷袭，他们决定将车队的长度变为原来的一半（向下取整），计划为：
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- 找出车队中 **物资之和最小** 两辆 **相邻** 马车，将它们车辆的物资整合为一辆。若存在多组物资之和相同的马车，则取编号最小的两辆马车进行整合；
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- 重复上述操作直到车队长度符合要求。
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- 找出车队中 **物资之和最小** 两辆 **相邻** 马车，将它们车辆的物资整合为一辆。若存在多组物资之和相同的马车，则取编号最小的两辆马车进行整合；
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- 重复上述操作直到车队长度符合要求。
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请返回车队长度符合要求后，物资的分布情况。
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**示例 1：**
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>输入：`supplies = [7,3,6,1,8]`
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> 输入：`supplies = [7,3,6,1,8]`
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>
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>输出：`[10,15]`
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> 输出：`[10,15]`
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>
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>解释：
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> 解释：
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> 第 1 次合并，符合条件的两辆马车为 6,1，合并后的车队为 [7,3,7,8]；
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> 第 2 次合并，符合条件的两辆马车为 (7,3) 和 (3,7)，取编号最小的 (7,3)，合并后的车队为 [10,7,8]；
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> 第 3 次合并，符合条件的两辆马车为 7,8，合并后的车队为 [10,15]；
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>返回 `[10,15]`
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> 返回 `[10,15]`
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**示例 2：**
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>输入：`supplies = [1,3,1,5]`
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> 输入：`supplies = [1,3,1,5]`
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>
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>输出：`[5,5]`
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> 输出：`[5,5]`
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**解释：**
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- `2 <= supplies.length <= 1000`
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- `1 <= supplies[i] <= 1000`
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- `2 <= supplies.length <= 1000`
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- `1 <= supplies[i] <= 1000`
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：模拟**
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根据题目描述，我们每次遍历 `supplies`，找到物资之和最小的两辆相邻马车，将它们车辆的物资整合为一辆，重复上述操作直到车队长度符合要求。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为 `supplies` 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def supplyWagon(self, supplies: List[int]) -> List[int]:
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for _ in range((len(supplies) + 1) >> 1):
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n = len(supplies)
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mi = inf
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k = 0
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for i in range(n - 1):
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x = supplies[i] + supplies[i + 1]
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if mi > x:
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mi = x
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k = i
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t = []
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i = 0
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while i < n:
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if i == k:
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t.append(mi)
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i += 2
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else:
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t.append(supplies[i])
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i += 1
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supplies = t
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return supplies
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] supplyWagon(int[] supplies) {
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for (int h = (supplies.length + 1) >> 1; h > 0; --h) {
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int n = supplies.length;
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int mi = 1 << 30;
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int k = 0;
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for (int i = 0; i < n - 1; ++i) {
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int x = supplies[i] + supplies[i + 1];
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if (mi > x) {
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mi = x;
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k = i;
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}
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}
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int[] t = new int[n - 1];
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for (int i = 0, j = 0; i < n; ++i, ++j) {
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if (i == k) {
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t[j] = mi;
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++i;
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} else {
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t[j] = supplies[i];
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}
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}
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supplies = t;
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}
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return supplies;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> supplyWagon(vector<int>& supplies) {
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for (int h = (supplies.size() + 1) >> 1; h; --h) {
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int n = supplies.size();
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int mi = 1 << 30;
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int k = 0;
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for (int i = 0; i < n - 1; ++i) {
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int x = supplies[i] + supplies[i + 1];
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if (mi > x) {
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mi = x;
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k = i;
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}
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}
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vector<int> t(n - 1);
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for (int i = 0, j = 0; i < n; ++i, ++j) {
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if (i == k) {
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t[j] = mi;
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++i;
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} else {
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t[j] = supplies[i];
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}
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}
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supplies = move(t);
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}
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return supplies;
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}
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};
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```
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### **Go**
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```go
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func supplyWagon(supplies []int) []int {
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for h := (len(supplies) + 1) >> 1; h > 0; h-- {
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n := len(supplies)
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mi := 1 << 30
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k := 0
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for i := 0; i < n-1; i++ {
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x := supplies[i] + supplies[i+1]
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if mi > x {
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mi = x
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k = i
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}
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}
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t := make([]int, n-1)
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for i, j := 0, 0; i < n; i, j = i+1, j+1 {
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if i == k {
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t[j] = mi
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i++
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} else {
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t[j] = supplies[i]
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}
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}
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supplies = t
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}
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return supplies
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}
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```
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### **TypeScript**
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```ts
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function supplyWagon(supplies: number[]): number[] {
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for (let h = (supplies.length + 1) >> 1; h > 0; --h) {
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const n = supplies.length;
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let mi = 1 << 30;
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let k = 0;
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for (let i = 0; i < n - 1; ++i) {
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const x = supplies[i] + supplies[i + 1];
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if (mi > x) {
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mi = x;
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k = i;
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}
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}
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const t: number[] = new Array(n - 1);
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for (let i = 0, j = 0; i < n; ++i, ++j) {
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if (i === k) {
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t[j] = mi;
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++i;
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} else {
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t[j] = supplies[i];
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}
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}
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supplies = t;
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}
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return supplies;
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}
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```
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### **...**

‎lcp/LCP 72. 补给马车/Solution.cpp

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class Solution {
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public:
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vector<int> supplyWagon(vector<int>& supplies) {
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for (int h = (supplies.size() + 1) >> 1; h; --h) {
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int n = supplies.size();
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int mi = 1 << 30;
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int k = 0;
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for (int i = 0; i < n - 1; ++i) {
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int x = supplies[i] + supplies[i + 1];
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if (mi > x) {
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mi = x;
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k = i;
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}
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}
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vector<int> t(n - 1);
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for (int i = 0, j = 0; i < n; ++i, ++j) {
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if (i == k) {
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t[j] = mi;
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++i;
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} else {
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t[j] = supplies[i];
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}
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}
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supplies = move(t);
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}
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return supplies;
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}
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};

‎lcp/LCP 72. 补给马车/Solution.go

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func supplyWagon(supplies []int) []int {
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for h := (len(supplies) + 1) >> 1; h > 0; h-- {
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n := len(supplies)
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mi := 1 << 30
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k := 0
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for i := 0; i < n-1; i++ {
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x := supplies[i] + supplies[i+1]
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if mi > x {
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mi = x
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k = i
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}
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}
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t := make([]int, n-1)
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for i, j := 0, 0; i < n; i, j = i+1, j+1 {
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if i == k {
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t[j] = mi
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i++
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} else {
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t[j] = supplies[i]
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}
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}
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supplies = t
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}
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return supplies
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}

‎lcp/LCP 72. 补给马车/Solution.java

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class Solution {
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public int[] supplyWagon(int[] supplies) {
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for (int h = (supplies.length + 1) >> 1; h > 0; --h) {
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int n = supplies.length;
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int mi = 1 << 30;
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int k = 0;
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for (int i = 0; i < n - 1; ++i) {
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int x = supplies[i] + supplies[i + 1];
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if (mi > x) {
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mi = x;
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k = i;
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}
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}
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int[] t = new int[n - 1];
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for (int i = 0, j = 0; i < n; ++i, ++j) {
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if (i == k) {
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t[j] = mi;
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++i;
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} else {
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t[j] = supplies[i];
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}
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}
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supplies = t;
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}
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return supplies;
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}
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}

‎lcp/LCP 72. 补给马车/Solution.py

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class Solution:
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def supplyWagon(self, supplies: List[int]) -> List[int]:
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for _ in range((len(supplies) + 1) >> 1):
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n = len(supplies)
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mi = inf
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k = 0
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for i in range(n - 1):
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x = supplies[i] + supplies[i + 1]
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if mi > x:
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mi = x
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k = i
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t = []
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i = 0
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while i < n:
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if i == k:
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t.append(mi)
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i += 2
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else:
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t.append(supplies[i])
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i += 1
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supplies = t
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return supplies

‎lcp/LCP 72. 补给马车/Solution.ts

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function supplyWagon(supplies: number[]): number[] {
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for (let h = (supplies.length + 1) >> 1; h > 0; --h) {
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const n = supplies.length;
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let mi = 1 << 30;
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let k = 0;
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for (let i = 0; i < n - 1; ++i) {
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const x = supplies[i] + supplies[i + 1];
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if (mi > x) {
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mi = x;
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k = i;
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}
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}
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const t: number[] = new Array(n - 1);
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for (let i = 0, j = 0; i < n; ++i, ++j) {
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if (i === k) {
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t[j] = mi;
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++i;
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} else {
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t[j] = supplies[i];
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}
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}
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supplies = t;
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}
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return supplies;
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}

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