1后没有0的最长子序列

给定一个二进制数组，求最长子序列的长度，使得 1 之后没有 0。

例子：

Input : 1 1 0 1
Output : 3
Explanation : 
If we remove 0 from the array, then no
zero comes right after one (satisfying 
the condition) and the maximum game 
left are 3 (i.e. 1 1 1)

Input : 0
Output : 1
Explanation : 
Since he wants to save maximum game in
the array. He doesn't remove any game.

让我们找出这个序列中有多少个零，然后取最后一个零之后的所有零。在每一步中，从序列的开头取下一个零并计算其后的零。用最大值更新答案。
您可以预先计算后缀上的个数。
例如0 1 0 0 1 1 1
计算后缀后，数组变为：
0 4 0 0 3 2 1
从头到尾移动，每次在数组中找到零时，将 numberofzeros 增加 1。如果 array[index] 不为零，则 res = max(res, numberofzeros + 该索引处数组的值)。
然后在循环之后： res = max(res, numberofzeros)

C++

// CPP program to find longest subsequence
// such that there is no 0 after 1.
#include 
using namespace std;
 
int maxSubseq(int vec[], int n) {   
         
    // Store the count of number of ones
    // from right to left in the array
    int suffix = 0;
    for (int i = n - 1; i >= 0; i--)
    {
        if (vec[i] == 1)
        {
            suffix++;
            vec[i] = suffix;
        }
    }
     
    // Traverse from left to right, keep count
    // of 0s and for every 0, check number of
    // 1s after it. Update the result if needed.
    int res = 0;
    int zero = 0;   
    for (int i = 0; i < n; i++)
    {
        if (vec[i] == 0)
            zero++;
     
        // Checking the maximum size
        if (vec[i] > 0)
            res = max(res, zero + vec[i]);
    }
     
    // Checking the maximum size
    return max(res, zero);
}
 
// Driver Code
int main()
{
    int input[] = { 0, 1, 0, 0, 1, 0 };
    int n = sizeof(input) / sizeof(input[0]);   
    cout << maxSubseq(input, n);
    return 0;
}

Java

// java program to find longest subsequence
// such that there is no 0 after 1.
import java.io.*;
 
public class GFG {
 
    static int maxSubseq(int []vec, int n)
    {
             
        // Store the count of number of
        // ones from right to left in
        // the array
        int suffix = 0;
         
        for (int i = n - 1; i >= 0; i--)
        {
            if (vec[i] == 1)
            {
                suffix++;
                vec[i] = suffix;
            }
        }
         
        // Traverse from left to right, keep
        // count of 0s and for every 0, check
        // number of 1s after it. Update the
        // result if needed.
        int res = 0;
        int zero = 0;
         
        for (int i = 0; i < n; i++)
        {
            if (vec[i] == 0)
                zero++;
         
            // Checking the maximum size
            if (vec[i] > 0)
                res = Math.max(res, zero + vec[i]);
        }
         
        // Checking the maximum size
        return Math.max(res, zero);
    }
     
    // Driver Code
    static public void main (String[] args)
    {
         
        int []input = { 0, 1, 0, 0, 1, 0 };
        int n = input.length;
         
        System.out.println(maxSubseq(input, n));
    }
}
 
// This code is contributed by vt_m.

Python3

# Python 3 program to find longest subsequence
# such that there is no 0 after 1.
 
def maxSubseq(vec, n):
    # Store the count of number of ones
    # from right to left in the array
    suffix = 0
    i = n-1
    while(i >= 0):
        if (vec[i] == 1):
            suffix += 1
            vec[i] = suffix
        i -= 1
             
    # Traverse from left to right, keep count
    # of 0s and for every 0, check number of
    # 1s after it. Update the result if needed.
    res = 0
    zero = 0
    for i in range(0,n,1):
        if (vec[i] == 0):
            zero += 1
     
        # Checking the maximum size
        if (vec[i] > 0):
            res = max(res, zero + vec[i])
     
    # Checking the maximum size
    return max(res, zero)
 
# Driver code
 if __name__ == '__main__':
    input = [0, 1, 0, 0, 1, 0]
    n = len(input)
    print(maxSubseq(input, n))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find longest subsequence
// such that there is no 0 after 1.
using System;
 
public class GFG {
 
    static int maxSubseq(int []vec, int n)
    {
             
        // Store the count of number of
        // ones from right to left in
        // the array
        int suffix = 0;
         
        for (int i = n - 1; i >= 0; i--)
        {
            if (vec[i] == 1)
            {
                suffix++;
                vec[i] = suffix;
            }
        }
         
        // Traverse from left to right, keep
        // count of 0s and for every 0, check
        // number of 1s after it. Update the
        // result if needed.
        int res = 0;
        int zero = 0;
         
        for (int i = 0; i < n; i++)
        {
            if (vec[i] == 0)
                zero++;
         
            // Checking the maximum size
            if (vec[i] > 0)
                res = Math.Max(res, zero + vec[i]);
        }
         
        // Checking the maximum size
        return Math.Max(res, zero);
    }
     
    // Driver Code
 
    static public void Main ()
    {
         
        int []input = { 0, 1, 0, 0, 1, 0 };
        int n = input.Length;
         
        Console.WriteLine(maxSubseq(input, n));
    }
}
 
// This code is contributed by vt_m.

PHP

= 0; $i--)
    {
        if ($vec[$i] == 1)
        {
            $suffix++;
            $vec[$i] = $suffix;
        }
    }
     
    // Traverse from left to
    // right, keep count of
    // 0s and for every 0,
    // check number of 1s after
    // it. Update the result if
    // needed.
    $res = 0;
    $zero = 0;
    for ($i = 0; $i < $n; $i++)
    {
        if ($vec[$i] == 0)
            $zero++;
     
        // Checking the
        // maximum size
        if ($vec[$i] > 0)
            $res = max($res, $zero +
                             $vec[$i]);
    }
     
    // Checking the
    // maximum size
    return max($res, $zero);
}
 
// Driver Code
$input = array(0, 1, 0, 0, 1, 0);
$n = count($input);
echo maxSubseq($input, $n);
 
// This code is contributed
// by anuj_67.
?>

Javascript

输出：