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feat: add solutions to lc problem: No.1478
No.1478.Allocate Mailboxes
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‎solution/1400-1499/1478.Allocate Mailboxes/README.md

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Original file line numberDiff line numberDiff line change
@@ -62,22 +62,149 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[i][j]$ 表示前 $i+1$ 栋房子，安排了 $j$ 个邮筒时，每栋房子与离它最近的邮筒之间的距离的最小总和。初始时 $f[i][j]=\infty$，答案即为 $f[n-1][k]$。
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我们可以枚举第 $j-1$ 个邮筒“管辖”的最后一栋房子 $p$，即 $0 \leq p \leq i-1$，那么第 $j$ 个邮筒“管辖”的房子就是 $[p+1,..i]$，我们记 $g[i][j]$ 表示给房子 $[i,..j]$ 安排一个邮筒的最小总和，那么有状态转移方程：
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$$
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f[i][j] = \min_{0 \leq p \leq i-1} \{f[p][j-1] + g[p+1][i]\}
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$$
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其中 $g[i][j]$ 的计算方法如下：
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$$
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]
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$$
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时间复杂度 $O(n^2 \times k)$，空间复杂度 $O(n^2)$。其中 $n$ 为房子的数量。
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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 minDistance(self, houses: List[int], k: int) -> int:
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houses.sort()
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n = len(houses)
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g = [[0] * n for _ in range(n)]
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]
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f = [[inf] * (k + 1) for _ in range(n)]
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for i in range(n):
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f[i][1] = g[0][i]
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for j in range(2, min(k + 1, i + 2)):
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for p in range(i):
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f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i])
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return f[-1][k]
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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 minDistance(int[] houses, int k) {
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Arrays.sort(houses);
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int n = houses.length;
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int[][] g = new int[n][n];
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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int[][] f = new int[n][k + 1];
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final int inf = 1 << 30;
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for (int[] e : f) {
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Arrays.fill(e, inf);
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}
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for (int i = 0; i < n; ++i) {
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f[i][1] = g[0][i];
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for (int j = 2; j <= k && j <= i + 1; ++j) {
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for (int p = 0; p < i; ++p) {
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f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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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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int minDistance(vector<int>& houses, int k) {
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int n = houses.size();
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sort(houses.begin(), houses.end());
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int g[n][n];
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memset(g, 0, sizeof(g));
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for (int i = n - 2; ~i; --i) {
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for (int j = i + 1; j < n; ++j) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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int f[n][k + 1];
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memset(f, 0x3f, sizeof(f));
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for (int i = 0; i < n; ++i) {
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f[i][1] = g[0][i];
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for (int j = 1; j <= k && j <= i + 1; ++j) {
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for (int p = 0; p < i; ++p) {
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f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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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 minDistance(houses []int, k int) int {
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sort.Ints(houses)
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n := len(houses)
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g := make([][]int, n)
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f := make([][]int, n)
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const inf = 1 << 30
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for i := range g {
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g[i] = make([]int, n)
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f[i] = make([]int, k+1)
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for j := range f[i] {
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f[i][j] = inf
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}
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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g[i][j] = g[i+1][j-1] + houses[j] - houses[i]
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}
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}
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for i := 0; i < n; i++ {
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f[i][1] = g[0][i]
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for j := 2; j <= k && j <= i+1; j++ {
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for p := 0; p < i; p++ {
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f[i][j] = min(f[i][j], f[p][j-1]+g[p+1][i])
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}
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}
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}
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return f[n-1][k]
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1478.Allocate Mailboxes/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -45,13 +45,122 @@ Minimum total distance from each houses to nearest mailboxes is |2-3| + |3-3| +
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### **Python3**
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```python
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class Solution:
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def minDistance(self, houses: List[int], k: int) -> int:
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houses.sort()
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n = len(houses)
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g = [[0] * n for _ in range(n)]
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]
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f = [[inf] * (k + 1) for _ in range(n)]
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for i in range(n):
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f[i][1] = g[0][i]
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for j in range(2, min(k + 1, i + 2)):
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for p in range(i):
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f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i])
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return f[-1][k]
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```
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### **Java**
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```java
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class Solution {
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public int minDistance(int[] houses, int k) {
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Arrays.sort(houses);
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int n = houses.length;
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int[][] g = new int[n][n];
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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int[][] f = new int[n][k + 1];
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final int inf = 1 << 30;
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for (int[] e : f) {
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Arrays.fill(e, inf);
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}
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for (int i = 0; i < n; ++i) {
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f[i][1] = g[0][i];
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for (int j = 2; j <= k && j <= i + 1; ++j) {
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for (int p = 0; p < i; ++p) {
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f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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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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int minDistance(vector<int>& houses, int k) {
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int n = houses.size();
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sort(houses.begin(), houses.end());
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int g[n][n];
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memset(g, 0, sizeof(g));
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for (int i = n - 2; ~i; --i) {
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for (int j = i + 1; j < n; ++j) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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int f[n][k + 1];
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memset(f, 0x3f, sizeof(f));
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for (int i = 0; i < n; ++i) {
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f[i][1] = g[0][i];
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for (int j = 1; j <= k && j <= i + 1; ++j) {
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for (int p = 0; p < i; ++p) {
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f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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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 minDistance(houses []int, k int) int {
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sort.Ints(houses)
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n := len(houses)
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g := make([][]int, n)
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f := make([][]int, n)
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const inf = 1 << 30
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for i := range g {
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g[i] = make([]int, n)
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f[i] = make([]int, k+1)
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for j := range f[i] {
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f[i][j] = inf
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}
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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g[i][j] = g[i+1][j-1] + houses[j] - houses[i]
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}
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}
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for i := 0; i < n; i++ {
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f[i][1] = g[0][i]
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for j := 2; j <= k && j <= i+1; j++ {
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for p := 0; p < i; p++ {
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f[i][j] = min(f[i][j], f[p][j-1]+g[p+1][i])
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}
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}
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}
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return f[n-1][k]
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1478.Allocate Mailboxes/Solution.cpp

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class Solution {
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public:
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int minDistance(vector<int>& houses, int k) {
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int n = houses.size();
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sort(houses.begin(), houses.end());
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int g[n][n];
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memset(g, 0, sizeof(g));
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for (int i = n - 2; ~i; --i) {
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for (int j = i + 1; j < n; ++j) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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int f[n][k + 1];
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memset(f, 0x3f, sizeof(f));
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for (int i = 0; i < n; ++i) {
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f[i][1] = g[0][i];
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for (int j = 1; j <= k && j <= i + 1; ++j) {
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for (int p = 0; p < i; ++p) {
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f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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}
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};

‎solution/1400-1499/1478.Allocate Mailboxes/Solution.go

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func minDistance(houses []int, k int) int {
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sort.Ints(houses)
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n := len(houses)
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g := make([][]int, n)
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f := make([][]int, n)
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const inf = 1 << 30
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for i := range g {
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g[i] = make([]int, n)
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f[i] = make([]int, k+1)
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for j := range f[i] {
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f[i][j] = inf
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}
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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g[i][j] = g[i+1][j-1] + houses[j] - houses[i]
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}
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}
19+
for i := 0; i < n; i++ {
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f[i][1] = g[0][i]
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for j := 2; j <= k && j <= i+1; j++ {
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for p := 0; p < i; p++ {
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f[i][j] = min(f[i][j], f[p][j-1]+g[p+1][i])
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}
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}
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}
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return f[n-1][k]
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}
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func min(a, b int) int {
31+
if a < b {
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return a
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}
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return b
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}

‎solution/1400-1499/1478.Allocate Mailboxes/Solution.java

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class Solution {
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public int minDistance(int[] houses, int k) {
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Arrays.sort(houses);
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int n = houses.length;
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int[][] g = new int[n][n];
6+
for (int i = n - 2; i >= 0; --i) {
7+
for (int j = i + 1; j < n; ++j) {
8+
g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
11+
int[][] f = new int[n][k + 1];
12+
final int inf = 1 << 30;
13+
for (int[] e : f) {
14+
Arrays.fill(e, inf);
15+
}
16+
for (int i = 0; i < n; ++i) {
17+
f[i][1] = g[0][i];
18+
for (int j = 2; j <= k && j <= i + 1; ++j) {
19+
for (int p = 0; p < i; ++p) {
20+
f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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}
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}

‎solution/1400-1499/1478.Allocate Mailboxes/Solution.py

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class Solution:
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def minDistance(self, houses: List[int], k: int) -> int:
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houses.sort()
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n = len(houses)
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g = [[0] * n for _ in range(n)]
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]
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f = [[inf] * (k + 1) for _ in range(n)]
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for i in range(n):
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f[i][1] = g[0][i]
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for j in range(2, min(k + 1, i + 2)):
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for p in range(i):
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f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i])
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return f[-1][k]

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