给定一个二进制字符串s ，任务是找到可以分成三个子字符串的最长子序列的长度，以使第一个和第三个子字符串为空或填充为1，而中间的子字符串为空或填充为0 。
例子：
Input: s = “1001”
Output: 4
Explanation:
The entire string can be divided into the desired three parts: “1”, “00”, “1”.
Input: s = “010”
Output: 2
Explanation:
The subsequence “00”, “01” and “10” can be split into desired three parts {“”, “00”, “”}, {“”, “0”, “1”} and {“1”, “0”, “”}
方法：
要解决该问题，我们需要执行以下步骤：
- 首先，预先计算并存储在前缀数组中，分别出现“ 1”和“ 0” 。
- 初始化两个整数i和j ，其中i是第一个字符串和第二个字符串之间的分隔点，而j是第二个字符串和第三个字符串之间的分隔点。
- 迭代i和j的所有可能值(0 <= i
下面是上述方法的实现：
C++
// C++ Program to find the
// longest subsequence possible
// that starts and ends with 1
// and filled with 0 in the middle
#include
using namespace std;
int longestSubseq(string s, int length)
{
// Prefix array to store the
// occurences of '1' and '0'
int ones[length + 1], zeroes[length + 1];
// Initialise prefix arrays with 0
memset(ones, 0, sizeof(ones));
memset(zeroes, 0, sizeof(zeroes));
// Iterate over the length of the string
for (int i = 0; i < length; i++) {
// If current character is '1'
if (s[i] == '1') {
ones[i + 1] = ones[i] + 1;
zeroes[i + 1] = zeroes[i];
}
// If current character is '0'
else {
zeroes[i + 1] = zeroes[i] + 1;
ones[i + 1] = ones[i];
}
}
int answer = INT_MIN;
int x = 0;
for (int i = 0; i <= length; i++) {
for (int j = i; j <= length; j++) {
// Add '1' available for
// the first string
x += ones[i];
// Add '0' available for
// the second string
x += (zeroes[j] - zeroes[i]);
// Add '1' available for
// the third string
x += (ones[length] - ones[j]);
// Update answer
answer = max(answer, x);
x = 0;
}
}
// Print the final result
cout << answer << endl;
}
// Driver Code
int main()
{
string s = "10010010111100101";
int length = s.length();
longestSubseq(s, length);
return 0;
} Java
// Java program to find the
// longest subsequence possible
// that starts and ends with 1
// and filled with 0 in the middle
import java.io.*;
class GFG{
static void longestSubseq(String s,
int length)
{
// Prefix array to store the
// occurences of '1' and '0'
int[] ones = new int[length + 1];
int[] zeroes = new int[length + 1];
// Iterate over the length of
// the string
for(int i = 0; i < length; i++)
{
// If current character is '1'
if (s.charAt(i) == '1')
{
ones[i + 1] = ones[i] + 1;
zeroes[i + 1] = zeroes[i];
}
// If current character is '0'
else
{
zeroes[i + 1] = zeroes[i] + 1;
ones[i + 1] = ones[i];
}
}
int answer = Integer.MIN_VALUE;
int x = 0;
for(int i = 0; i <= length; i++)
{
for(int j = i; j <= length; j++)
{
// Add '1' available for
// the first string
x += ones[i];
// Add '0' available for
// the second string
x += (zeroes[j] - zeroes[i]);
// Add '1' available for
// the third string
x += (ones[length] - ones[j]);
// Update answer
answer = Math.max(answer, x);
x = 0;
}
}
// Print the final result
System.out.println(answer);
}
// Driver code
public static void main(String[] args)
{
String s = "10010010111100101";
int length = s.length();
longestSubseq(s, length);
}
}
// This code is contributed by offbeatPython3
# Python3 program to find the
# longest subsequence possible
# that starts and ends with 1
# and filled with 0 in the middle
import sys
def longestSubseq(s, length):
# Prefix array to store the
# occurences of '1' and '0'
# Initialise prefix arrays with 0
ones = [0 for i in range(length + 1)]
zeroes = [0 for i in range(length + 1)]
# Iterate over the length of the string
for i in range(length):
# If current character is '1'
if(s[i] == '1'):
ones[i + 1] = ones[i] + 1
zeroes[i + 1] = zeroes[i]
# If current character is '0'
else:
zeroes[i + 1] = zeroes[i] + 1
ones[i + 1] = ones[i]
answer = -sys.maxsize - 1
x = 0
for i in range(length + 1):
for j in range(i, length + 1):
# Add '1' available for
# the first string
x += ones[i]
# Add '0' available for
# the second string
x += (zeroes[j] - zeroes[i])
# Add '1' available for
# the third string
x += (ones[length] - ones[j])
# Update answer
answer = max(answer, x)
x = 0
# Print the final result
print(answer)
# Driver Code
S = "10010010111100101"
length = len(S)
longestSubseq(S, length)
# This code is contributed by avanitrachhadiya2155C#
// C# program to find the
// longest subsequence possible
// that starts and ends with 1
// and filled with 0 in the middle
using System;
class GFG{
static void longestSubseq(String s,
int length)
{
// Prefix array to store the
// occurences of '1' and '0'
int[] ones = new int[length + 1];
int[] zeroes = new int[length + 1];
// Iterate over the length of
// the string
for(int i = 0; i < length; i++)
{
// If current character is '1'
if (s[i] == '1')
{
ones[i + 1] = ones[i] + 1;
zeroes[i + 1] = zeroes[i];
}
// If current character is '0'
else
{
zeroes[i + 1] = zeroes[i] + 1;
ones[i + 1] = ones[i];
}
}
int answer = int.MinValue;
int x = 0;
for(int i = 0; i <= length; i++)
{
for(int j = i; j <= length; j++)
{
// Add '1' available for
// the first string
x += ones[i];
// Add '0' available for
// the second string
x += (zeroes[j] - zeroes[i]);
// Add '1' available for
// the third string
x += (ones[length] - ones[j]);
// Update answer
answer = Math.Max(answer, x);
x = 0;
}
}
// Print the readonly result
Console.WriteLine(answer);
}
// Driver code
public static void Main(String[] args)
{
String s = "10010010111100101";
int length = s.Length;
longestSubseq(s, length);
}
}
// This code is contributed by Amit Katiyar输出：
12
时间复杂度： O(N 2 )
辅助空间： O(N)