java 解法

好的，下面是这个问题的一个Java实现:

public boolean canPlaceFlowers(int[] flowerbed, int n) {
    int count = 0;
    int m = flowerbed.length;
    for (int i = 0; i < m; i++) {
        if (flowerbed[i] == 0) {
            if (i == 0 || flowerbed[i-1] == 0) {
                if (i == m-1 || flowerbed[i+1] == 0) {
                    flowerbed[i] = 1;
                    count++;
                }
            }
        }
        if (count >= n) {
            return true;
        }
    }
    return false;
}

这个算法的时间复杂度为 O(n)，其中 n 为花坛的长度。

相关问题

hdu4310 java解法

以下是hdu4310的Java解法：

import java.util.*;
import java.io.*;

public class Main {
    static int MAXN = 100010;
    static int MAXM = 200010;
    static int INF = 0x3f3f3f3f;

    static int n, m, s, t, cnt;
    static int[] head = new int[MAXN];
    static int[] dis = new int[MAXN];
    static boolean[] vis = new boolean[MAXN];
    static int[] pre = new int[MAXN];
    static int[] cur = new int[MAXN];

    static class Edge {
        int to, next, cap, flow, cost;

        public Edge(int to, int next, int cap, int flow, int cost) {
            this.to = to;
            this.next = next;
            this.cap = cap;
            this.flow = flow;
            this.cost = cost;
        }
    }

    static Edge[] edge = new Edge[MAXM];

    static void addEdge(int u, int v, int cap, int flow, int cost) {
        edge[cnt] = new Edge(v, head[u], cap, flow, cost);
        head[u] = cnt++;        edge[cnt] = new Edge(u, head[v], 0, 0, -cost);
        head[v] = cnt++;
    }

    static boolean spfa() {
        Arrays.fill(dis, INF);
        Arrays.fill(vis, false);
        Queue<Integer> q = new LinkedList<>();
        q.offer(s);
        dis[s] = 0;
        vis[s] = true;
        while (!q.isEmpty()) {
            int u = q.poll();
            vis[u] = false;
            for (int i = head[u]; i != -1; i = edge[i].next) {
                int v = edge[i].to;
                if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost) {
                    dis[v] = dis[u] + edge[i].cost;
                    pre[v] = u;
                    cur[v] = i;
                    if (!vis[v]) {
                        vis[v] = true;
                        q.offer(v);
                    }
                }
            }
        }
        return dis[t] != INF;
    }

    static int[] MCMF(int s, int t) {
        int flow = 0, cost = 0;
        while (spfa()) {
            int f = INF;
            for (int u = t; u != s; u = pre[u]) {
                f = Math.min(f, edge[cur[u]].cap - edge[cur[u]].flow);
            }
            for (int u = t; u != s; u = pre[u]) {
                edge[cur[u]].flow += f;
                edge[cur[u] ^ 1].flow -= f;
                cost += edge[cur[u]].cost * f;
            }
            flow += f;
        }
        return new int[]{flow, cost};
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int T = in.nextInt();
        for (int cas = 1; cas <= T; cas++) {
            n = in.nextInt();
            m = in.nextInt();
            s = 0;
            t = n + m + 1;
            cnt = 0;
            Arrays.fill(head, -1);
            for (int i = 1; i <= n; i++) {
                int c = in.nextInt();
                addEdge(s, i, c, 0, 0);
            }
            for (int i = 1; i <= m; i++) {
                int c = in.nextInt();
                addEdge(i + n, t, c, 0, 0);
            }
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    int c = in.nextInt();
                    addEdge(i, j + n, INF, 0, c);
                }
            }
            int[] ans = MCMF(s, t);
            System.out.printf("Case #%d: %d\n", cas, ans[1]);
        }
    }
}

最小循环子数组 java解法

最小循环子数组是指在一个循环数组中，找到一个连续子数组，使该子数组的和最小。对于解决这个问题，可以使用Java编程语言来实现。

首先，我们需要定义一个函数来计算数组的和。该函数的输入参数为一个整数数组，返回值为数组的总和。下面是该函数的Java代码实现：

public int getSum(int[] nums) {
    int sum = 0;
    for (int num : nums) {
        sum += num;
    }
    return sum;
}

接下来，我们可以编写一个函数来找到最小循环子数组的和。该函数的输入参数为一个整数数组，返回值为最小循环子数组的和。下面是该函数的Java代码实现：

public int findMinSubarraySum(int[] nums) {
    int sum = getSum(nums);
    int currentSum = 0;
    int minSum = sum;

    for (int i = 0; i < nums.length; i++) {
        currentSum += nums[i];
        if (currentSum > sum) {
            currentSum = nums[i];
        }
        minSum = Math.min(minSum, currentSum);
    }

    return minSum;
}

在这段代码中，我们使用了一个变量currentSum来记录当前子数组的和，另一个变量minSum来记录最小循环子数组的和。在遍历数组过程中，如果当前子数组的和大于整个数组的总和，就将currentSum重置为当前元素的值。每次更新minSum时，都使用Math.min函数来比较两个和的大小，并将较小的值赋给minSum。

最后，我们可以调用findMinSubarraySum函数来找到最小循环子数组的和。下面是一个例子：

int[] nums = {4, 3, -2, -1, -3, 4, 5};
int minSubarraySum = findMinSubarraySum(nums);
System.out.println("最小循环子数组的和为：" + minSubarraySum);

以上就是最小循环子数组问题的Java解法，通过以上的代码实现，可以计算出最小循环子数组的和。