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feat: add solutions to lcp problem: No.39 (doocs#1533)
No.39.无人机方阵
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‎lcp/LCP 39. 无人机方阵/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表计数**
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我们可以用哈希表 $cnt$ 统计 $source$ 和 $target$ 中每个数字出现的次数之差。对于 $cnt$ 中的每个数字 $x$，如果 $x$ 的出现次数为正数，那么说明 $x$ 在 $target$ 中出现的次数多，我们需要将 $x$ 出现的次数减少到 $0$。因此，我们只需要累加所有出现次数为正数的数字的出现次数之和，即为答案。
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时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是 $source$ 和 $target$ 的行数和列数。
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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 minimumSwitchingTimes(
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self, source: List[List[int]], target: List[List[int]]
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) -> int:
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cnt = Counter()
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for row in source:
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for x in row:
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cnt[x] += 1
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for row in target:
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for x in row:
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cnt[x] -= 1
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return sum(abs(x) for x in cnt.values()) // 2
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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 minimumSwitchingTimes(int[][] source, int[][] target) {
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Map<Integer, Integer> cnt = new HashMap<>();
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for (int[] row : source) {
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for (int x : row) {
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cnt.merge(x, 1, Integer::sum);
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}
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}
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for (int[] row : target) {
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for (int x : row) {
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cnt.merge(x, -1, Integer::sum);
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}
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}
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int ans = 0;
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for (int v : cnt.values()) {
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ans += Math.abs(v);
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}
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return ans / 2;
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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 minimumSwitchingTimes(vector<vector<int>>& source, vector<vector<int>>& target) {
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unordered_map<int, int> cnt;
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for (auto& row : source) {
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for (int& x : row) {
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++cnt[x];
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}
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}
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for (auto& row : target) {
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for (int& x : row) {
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--cnt[x];
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}
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}
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int ans = 0;
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for (auto& [_, v] : cnt) {
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ans += abs(v);
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}
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return ans / 2;
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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 minimumSwitchingTimes(source [][]int, target [][]int) (ans int) {
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cnt := map[int]int{}
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for _, row := range source {
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for _, x := range row {
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cnt[x]++
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}
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}
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for _, row := range target {
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for _, x := range row {
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cnt[x]--
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}
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}
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for _, v := range cnt {
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if v > 0 {
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ans += v
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function minimumSwitchingTimes(source: number[][], target: number[][]): number {
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const cnt: Map<number, number> = new Map();
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for (const row of source) {
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for (const x of row) {
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cnt.set(x, (cnt.get(x) || 0) + 1);
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}
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}
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for (const row of target) {
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for (const x of row) {
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cnt.set(x, (cnt.get(x) || 0) - 1);
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}
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}
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let ans = 0;
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for (const [_, v] of cnt) {
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ans += Math.abs(v);
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}
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return Math.floor(ans / 2);
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}
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```
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### **...**

‎lcp/LCP 39. 无人机方阵/Solution.cpp

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class Solution {
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public:
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int minimumSwitchingTimes(vector<vector<int>>& source, vector<vector<int>>& target) {
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unordered_map<int, int> cnt;
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for (auto& row : source) {
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for (int& x : row) {
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++cnt[x];
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}
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}
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for (auto& row : target) {
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for (int& x : row) {
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--cnt[x];
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}
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}
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int ans = 0;
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for (auto& [_, v] : cnt) {
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ans += abs(v);
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}
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return ans / 2;
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}
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};

‎lcp/LCP 39. 无人机方阵/Solution.go

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func minimumSwitchingTimes(source [][]int, target [][]int) (ans int) {
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cnt := map[int]int{}
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for _, row := range source {
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for _, x := range row {
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cnt[x]++
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}
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}
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for _, row := range target {
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for _, x := range row {
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cnt[x]--
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}
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}
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for _, v := range cnt {
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if v > 0 {
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ans += v
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}
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}
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return
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}

‎lcp/LCP 39. 无人机方阵/Solution.java

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class Solution {
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public int minimumSwitchingTimes(int[][] source, int[][] target) {
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Map<Integer, Integer> cnt = new HashMap<>();
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for (int[] row : source) {
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for (int x : row) {
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cnt.merge(x, 1, Integer::sum);
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}
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}
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for (int[] row : target) {
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for (int x : row) {
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cnt.merge(x, -1, Integer::sum);
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}
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}
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int ans = 0;
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for (int v : cnt.values()) {
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ans += Math.abs(v);
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}
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return ans / 2;
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}
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}

‎lcp/LCP 39. 无人机方阵/Solution.py

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class Solution:
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def minimumSwitchingTimes(
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self, source: List[List[int]], target: List[List[int]]
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) -> int:
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cnt = Counter()
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for row in source:
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for x in row:
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cnt[x] += 1
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for row in target:
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for x in row:
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cnt[x] -= 1
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return sum(abs(x) for x in cnt.values()) // 2

‎lcp/LCP 39. 无人机方阵/Solution.ts

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function minimumSwitchingTimes(source: number[][], target: number[][]): number {
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const cnt: Map<number, number> = new Map();
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for (const row of source) {
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for (const x of row) {
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cnt.set(x, (cnt.get(x) || 0) + 1);
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}
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}
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for (const row of target) {
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for (const x of row) {
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cnt.set(x, (cnt.get(x) || 0) - 1);
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}
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}
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let ans = 0;
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for (const [_, v] of cnt) {
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ans += Math.abs(v);
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}
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return Math.floor(ans / 2);
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}

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