题目描述:
在一个大型体育场内举办了一场大型活动，由于疫情防控的需要，要求每位观众的必须间隔至少一个空位才允许落座。现在给出一排观众座位 分布图，座位中存在已落座的观众，请计算出，在不移动现有观众座位的情况下，最多还能坐下多少名观众。输入描述:
一个数组，用来标识某一排座位中，每个座位是否已经坐人。0表示该座位没有坐人，1表示该座位已经坐人。输出描述:
整数，在不移动现有观众座位的情况下，最多还能坐下多少名观众。补充说明:
1<=数组长度<=10000
示例1
输入:
10001
输出:
1
说明:
示例2
输入:
0101
输出:
0

Java解法：

public class MaxAdditionalSeats {

    public static int maxAdditionalSeats(int[] seats) {
        int count = 0; // 新观众的计数
        int n = seats.length;

        for (int i = 0; i < n; i++) {
            // 如果当前座位是空的
            if (seats[i] == 0) {
                // 检查左边是否空（或左边界）
                boolean leftEmpty = (i == 0) || (seats[i - 1] == 0);
                // 检查右边是否空（或右边界）
                boolean rightEmpty = (i == n - 1) || (seats[i + 1] == 0);

                // 如果左右都是空的，当前座位可以坐人
                if (leftEmpty && rightEmpty) {
                    count++; // 新增观众计数
                    i++; // 跳过下一个座位，因为不能相邻坐人
                }
            }
        }

        return count;
    }

    public static void main(String[] args) {
        int[] seats1 = {1, 0, 0, 0, 1, 0, 1};
        System.out.println(maxAdditionalSeats(seats1)); // 输出应为 1

        int[] seats2 = {0, 0, 0, 0, 0};
        System.out.println(maxAdditionalSeats(seats2)); // 输出应为 2

        int[] seats3 = {1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1};
        System.out.println(maxAdditionalSeats(seats3)); // 输出应为 2

        int[] seats4 = {1, 0, 0, 0, 0, 1};
        System.out.println(maxAdditionalSeats(seats4)); // 输出应为 1

        int[] seats5 = {0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1};
        System.out.println(maxAdditionalSeats(seats5)); // 输出应为 3
    }
}

C++解法：

#include <iostream>
#include <string>
#include <tuple>
#include <vector>
 
using namespace std;
 
int main() {
  string in;
  cin >> in;
  size_t len = in.length();
  int res = 0;
  if (len == 1) {
    if (in[0] == '0')
      cout << 1 << endl;
    else
      cout << 0 << endl;
    return 0;
  }
 
  for (int i = 0; i < len; i++) {
    auto curr_char = in[i];
    if (curr_char == '1') {
      if (i - 1 >= 0) {
        in[i - 1] = '_';
      }
      if (i + 1 < len) {
        in[i + 1] = '_';
      }
    }
 
    if (curr_char == '0') {
 
      if (i + 1 == len) {
        auto prev_char = in[i - 1];
        if (prev_char == '0' || prev_char == '_') {
          in[i] = '1';
          res += 1;
        }
      } else if (i - 1 == -1) {
        auto next_char = in[i + 1];
        if (next_char == '0' || next_char == '_') {
          in[i] = '1';
          res += 1;
        }
      } else {
        auto prev_char = in[i - 1];
        auto next_char = in[i + 1];
        if ((prev_char == '0' || prev_char == '_') &&
            (next_char == '0' || next_char == '_')) {
          in[i] = '1';
          res += 1;
        }
      }
    }
  }
 
  cout << res << endl;
 
  return 0;
}

Python解法：

while True:
    try:
        arr = input()
        if len(arr) <= 2:
            if '1' in arr:
                print(0)
                break
            else:
                print(1)
                break
        
        count0 = 1
        countN = 0
        for s in arr:
            if s == '0':
                count0 += 1
                if count0 == 3:
                    countN += 1
                    count0 = 1
            else:
                count0 = 0
        
       
        if arr[-1] != '1' and '1' in arr:
            temp = arr.split('1')[-1]
           
            if len(temp) % 2 == 0:
                countN += 1
        
        print(countN)
 
    except:
        break