/** * 获取 list 是否为空(true) * @return list 是否为空 */ boolean isEmpty();数组方法实现代码

Sure! Here is an example implementation of the `isEmpty()` method using an array:

public class MyList {
    private int[] array;

    public MyList() {
        array = new int[0];
    }

    public boolean isEmpty() {
        return array.length == 0;
    }
}

In this example, the `isEmpty()` method checks if the length of the array is 0, indicating that the list is empty. If the length is 0, it returns `true`; otherwise, it returns `false`.

相关问题

有一个数据库（rouble_personnel_pool），实体类为RoublePersonnelPool的字段参数类型有numbleId（序号），roubleUnit（隶属单位），roubleBumen（部门），roubleName（人员名称），roubleID（编号），而且都是string类型。以下是model层 @Data @Entity @Table(name = "rouble_personnel_pool") @EqualsAndHashCode(callSuper = true) public class RoublePersonnelPool extends BaseEntity { private static final long serialVersionUID = 1L; /** * 序号 */ @Column(name = "number_id") @ApiModelProperty("序号") @NotBlank(message = "序号不能为空") private String numberId; /** * 隶属单位 */ @Column(name = "rouble_unit") @ApiModelProperty("隶属单位") @NotBlank(message = "隶属单位不能为空") private String roubleUnit; /** * 部门 */ @Column(name = "rouble_bumen") @ApiModelProperty("部门") @NotBlank(message = "部门不能为空") private String roubleBumen; /** * 人员名称 */ @Column(name = "rouble_name") @ApiModelProperty("人员名称") @NotBlank(message = "人员名称不能为空") private String roubleName; /** * 编号 */ @Column(name = "rouble_ID") @ApiModelProperty("编号") @NotBlank(message = "编号不能为空") private String roubleID; } 帮我完成接口实现类和controller层，要求里面要有RoublePersonnelPoolRepository类并且extends BaseRepository<RoubleObjectPool,String>。 帮我写一套增删改查。

接口实现类和控制器层的代码如下：

RoublePersonnelPoolRepository.java:

@Repository
public interface RoublePersonnelPoolRepository extends BaseRepository<RoubleObjectPool, String> {
}

RoublePersonnelPoolServiceImpl.java:

@Service
public class RoublePersonnelPoolServiceImpl implements RoublePersonnelPoolService {

    private final RoublePersonnelPoolRepository roublePersonnelPoolRepository;

    @Autowired
    public RoublePersonnelPoolServiceImpl(RoublePersonnelPoolRepository roublePersonnelPoolRepository) {
        this.roublePersonnelPoolRepository = roublePersonnelPoolRepository;
    }

    @Override
    public RoublePersonnelPool getById(String id) {
        return roublePersonnelPoolRepository.findById(id).orElse(null);
    }

    @Override
    public List<RoublePersonnelPool> getAll() {
        return roublePersonnelPoolRepository.findAll();
    }

    @Override
    public RoublePersonnelPool create(RoublePersonnelPool roublePersonnelPool) {
        return roublePersonnelPoolRepository.save(roublePersonnelPool);
    }

    @Override
    public RoublePersonnelPool update(String id, RoublePersonnelPool roublePersonnelPool) {
        RoublePersonnelPool existingRoublePersonnelPool = getById(id);
        if (existingRoublePersonnelPool != null) {
            existingRoublePersonnelPool.setNumberId(roublePersonnelPool.getNumberId());
            existingRoublePersonnelPool.setRoubleUnit(roublePersonnelPool.getRoubleUnit());
            existingRoublePersonnelPool.setRoubleBumen(roublePersonnelPool.getRoubleBumen());
            existingRoublePersonnelPool.setRoubleName(roublePersonnelPool.getRoubleName());
            existingRoublePersonnelPool.setRoubleID(roublePersonnelPool.getRoubleID());
            return roublePersonnelPoolRepository.save(existingRoublePersonnelPool);
        }
        return null;
    }

    @Override
    public boolean delete(String id) {
        RoublePersonnelPool existingRoublePersonnelPool = getById(id);
        if (existingRoublePersonnelPool != null) {
            roublePersonnelPoolRepository.delete(existingRoublePersonnelPool);
            return true;
        }
        return false;
    }
}

RoublePersonnelPoolController.java:

@RestController
@RequestMapping("/rouble-personnel-pool")
public class RoublePersonnelPoolController {

    private final RoublePersonnelPoolService roublePersonnelPoolService;

    @Autowired
    public RoublePersonnelPoolController(RoublePersonnelPoolService roublePersonnelPoolService) {
        this.roublePersonnelPoolService = roublePersonnelPoolService;
    }

    @GetMapping("/{id}")
    public ResponseEntity<RoublePersonnelPool> getById(@PathVariable String id) {
        RoublePersonnelPool roublePersonnelPool = roublePersonnelPoolService.getById(id);
        if (roublePersonnelPool != null) {
            return ResponseEntity.ok(roublePersonnelPool);
        }
        return ResponseEntity.notFound().build();
    }

    @GetMapping
    public ResponseEntity<List<RoublePersonnelPool>> getAll() {
        List<RoublePersonnelPool> roublePersonnelPools = roublePersonnelPoolService.getAll();
        if (!roublePersonnelPools.isEmpty()) {
            return ResponseEntity.ok(roublePersonnelPools);
        }
        return ResponseEntity.noContent().build();
    }

    @PostMapping
    public ResponseEntity<RoublePersonnelPool> create(@RequestBody RoublePersonnelPool roublePersonnelPool) {
        RoublePersonnelPool createdRoublePersonnelPool = roublePersonnelPoolService.create(roublePersonnelPool);
        if (createdRoublePersonnelPool != null) {
            return ResponseEntity.ok(createdRoublePersonnelPool);
        }
        return ResponseEntity.badRequest().build();
    }

    @PutMapping("/{id}")
    public ResponseEntity<RoublePersonnelPool> update(@PathVariable String id, @RequestBody RoublePersonnelPool roublePersonnelPool) {
        RoublePersonnelPool updatedRoublePersonnelPool = roublePersonnelPoolService.update(id, roublePersonnelPool);
        if (updatedRoublePersonnelPool != null) {
            return ResponseEntity.ok(updatedRoublePersonnelPool);
        }
        return ResponseEntity.notFound().build();
    }

    @DeleteMapping("/{id}")
    public ResponseEntity<Void> delete(@PathVariable String id) {
        boolean deleted = roublePersonnelPoolService.delete(id);
        if (deleted) {
            return ResponseEntity.noContent().build();
        }
        return ResponseEntity.notFound().build();
    }
}

请注意，这只是一个基本的示例，你可能需要根据你的业务逻辑进行相应的修改和调整。

描述有两棵大小为n且以1为根的树，设为树1和树2，令根的深度为0，在树1上编号为x的点，在树2中编号为a[x]。定义点对(x,y)的价值为树1中x和y的最近公共祖先的深度+树2中a[x]和a[y]的最近公共祖先的深度。输出所有点对的价值的最大值。示例输入：3,[0,2,1,3],[0,0,1,1],[0,0,1,2]返回值：1备注：输入时给出n和三个数组a,b,c。a数组下标从0到n，意义如题,保证a[0]=0,a[1]~a[n]的数互不相同且1~n各只出现一次。b数组下标从0到n，为描述树1的数组，b[x]表示树1上x的父亲节点的编号，其中b[0]=b[1]=0.c数组下标从0到n，为描述树2的数组，c[x]表示树2上x的父亲节点的编号，其中c[0]=c[1]=0.输出一个整数，表示最大的价值。n<=100000，考虑到n很大的情况会使得运行效率变慢，请使用时间和空间复杂度较低的算法，并请按照以下模板进行实现： import java.util.*; public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param n int整型 * @param a int整型一维数组 * @param b int整型一维数组 * @param c int整型一维数组 * @return int整型 */ public int wwork (int n, int[] a, int[] b, int[] c) { // write code here } }

以下是Java实现的代码：

import java.util.*;

public class Solution {
    public int wwork(int n, int[] a, int[] b, int[] c) {
        List<Integer>[] tree1 = buildTree(b);
        List<Integer>[] tree2 = buildTree(c);

        // 记录每个节点在树1和树2中的编号
        int[] id1 = new int[n + 1];
        int[] id2 = new int[n + 1];
        int cnt = 0;
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(1);
        while (!queue.isEmpty()) {
            int u = queue.poll();
            id1[u] = ++cnt;
            for (int v : tree1[u]) {
                queue.offer(v);
            }
        }
        cnt = 0;
        queue.offer(1);
        while (!queue.isEmpty()) {
            int u = queue.poll();
            id2[u] = ++cnt;
            for (int v : tree2[u]) {
                queue.offer(v);
            }
        }

        // 预处理树1和树2的LCA
        int[][] lca1 = buildLca(tree1, id1);
        int[][] lca2 = buildLca(tree2, id2);

        // 计算所有点对的价值并取最大值
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                int lca1Depth = lca1[id1[i]][id1[j]];
                int lca2Depth = lca2[id2[a[i]]][id2[a[j]]];
                ans = Math.max(ans, lca1Depth + lca2Depth);
            }
        }

        return ans;
    }

    // 建立邻接表表示树
    private List<Integer>[] buildTree(int[] parent) {
        int n = parent.length;
        List<Integer>[] tree = new List[n];
        for (int i = 0; i < n; i++) {
            tree[i] = new ArrayList<>();
        }
        for (int i = 1; i < n; i++) {
            tree[parent[i]].add(i);
        }
        return tree;
    }

    // 建立LCA数组
    private int[][] buildLca(List<Integer>[] tree, int[] id) {
        int n = tree.length;
        int maxDepth = (int) Math.floor(Math.log(n) / Math.log(2));
        int[][] lca = new int[n][maxDepth + 1];
        boolean[] visited = new boolean[n];
        dfs(tree, id, lca, visited, 1, 0);
        for (int k = 1; k <= maxDepth; k++) {
            for (int i = 1; i < n; i++) {
                lca[i][k] = lca[lca[i][k - 1]][k - 1];
            }
        }
        return lca;
    }

    // DFS求LCA数组
    private void dfs(List<Integer>[] tree, int[] id, int[][] lca, boolean[] visited, int u, int depth) {
        visited[u] = true;
        lca[id[u]][0] = u;
        for (int v : tree[u]) {
            if (!visited[v]) {
                dfs(tree, id, lca, visited, v, depth + 1);
                lca[id[v]][0] = u;
            }
        }
    }
}

时间复杂度：$O(n\log n)$。

空间复杂度：$O(n)$。