给定一个二进制字符串S ，任务是通过反转偶数为1的子字符串将给定的字符串S转换为其词典序的最大形式。
例子：
Input: S = “10101”
Output: 11010
Explanation:
Reversing the substring {S[0], …, S[2]} modifies S to “10101”.
Reversing the substring {S[1], …, S[4]} modifies S to “11010”.
Now, any other substring cannot be chosen as each will either have odd number of 1s or will be lexicographically smaller than the previous string. Thus, 11010 is the lexicographically largest string that can be performed.
Input: S = “0101”
Output: 1010
方法：我们的想法是从第一个索引开始，遍历字符串，直到1秒的字符串中的总数变为偶数。一旦找到恰好为偶数1的索引，就将字符串从起始索引反向转换为结束索引。对每个索引重复此过程，以得到字典上最大的字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the lexicographically
// maximum string by reversing substrings
// having even numbers of 1s
void lexicographicallyMax(string s)
{
// Store size of string
int n = s.size();
// Traverse the string
for (int i = 0; i < n; i++) {
// Count the number of 1s
int count = 0;
// Stores the starting index
int beg = i;
// Stores the end index
int end = i;
// Increment count, when 1
// is encountered
if (s[i] == '1')
count++;
// Traverse the remaining string
for (int j = i + 1; j < n; j++) {
if (s[j] == '1')
count++;
if (count % 2 == 0
&& count != 0) {
end = j;
break;
}
}
// Reverse the string from
// starting and end index
reverse(s.begin() + beg,
s.begin() + end + 1);
}
// Printing the string
cout << s << "\n";
}
// Driver Code
int main()
{
string S = "0101";
lexicographicallyMax(S);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the lexicographically
// maximum string by reversing substrings
// having even numbers of 1s
static void lexicographicallyMax(String s)
{
// Store size of string
int n = s.length();
// Traverse the string
for(int i = 0; i < n; i++)
{
// Count the number of 1s
int count = 0;
// Stores the starting index
int beg = i;
// Stores the end index
int end = i;
// Increment count, when 1
// is encountered
if (s.charAt(i) == '1')
count++;
// Traverse the remaining string
for(int j = i + 1; j < n; j++)
{
if (s.charAt(j) == '1')
count++;
if (count % 2 == 0 &&
count != 0)
{
end = j;
break;
}
}
// Reverse the string from
// starting and end index
s = reverse(s, beg, end + 1);
}
// Printing the string
System.out.println(s);
}
static String reverse(String s, int beg, int end)
{
StringBuilder x = new StringBuilder("");
for(int i = 0; i < beg; i++)
x.append(s.charAt(i));
for(int i = end - 1; i >= beg; i--)
x.append(s.charAt(i));
for(int i = end; i < s.length(); i++)
x.append(s.charAt(i));
return x.toString();
}
// Driver Code
public static void main(String args[])
{
String S = "0101";
lexicographicallyMax(S);
}
}
// This code is contributed by jyoti369Python3
# Python3 program for the above approach
# Function to find the lexicographically
# maximum string by reversing substrings
# having even numbers of 1s
def lexicographicallyMax(s):
# Store size of string
n = len(s)
# Traverse the string
for i in range(n):
# Count the number of 1s
count = 0
# Stores the starting index
beg = i
# Stores the end index
end = i
# Increment count, when 1
# is encountered
if (s[i] == '1'):
count += 1
# Traverse the remaining string
for j in range(i + 1, n):
if (s[j] == '1'):
count += 1
if (count % 2 == 0 and count != 0):
end = j
break
# temp is for Reverse the string from
# starting and end index
temp = s[beg : end + 1]
temp = temp[::-1]
s = s[0 : beg] + temp + s[end + 1 :]
# Printing the string
print(s)
# Driver Code
S = "0101"
lexicographicallyMax(S)
# This code is contributed by avanitrachhadiya2155C#
// C# program for the above approach
using System;
using System.Text;
class GFG
{
// Function to find the lexicographically
// maximum string by reversing substrings
// having even numbers of 1s
static void lexicographicallyMax(String s)
{
// Store size of string
int n = s.Length;
// Traverse the string
for(int i = 0; i < n; i++)
{
// Count the number of 1s
int count = 0;
// Stores the starting index
int beg = i;
// Stores the end index
int end = i;
// Increment count, when 1
// is encountered
if (s[i] == '1')
count++;
// Traverse the remaining string
for(int j = i + 1; j < n; j++)
{
if (s[j] == '1')
count++;
if (count % 2 == 0 &&
count != 0)
{
end = j;
break;
}
}
// Reverse the string from
// starting and end index
s = reverse(s, beg, end + 1);
}
// Printing the string
Console.WriteLine(s);
}
static String reverse(String s, int beg, int end)
{
StringBuilder x = new StringBuilder("");
for(int i = 0; i < beg; i++)
x.Append(s[i]);
for(int i = end - 1; i >= beg; i--)
x.Append(s[i]);
for(int i = end; i < s.Length; i++)
x.Append(s[i]);
return x.ToString();
}
// Driver Code
public static void Main(String []args)
{
String S = "0101";
lexicographicallyMax(S);
}
}
// This code is contributed by Princi Singh输出：
1010时间复杂度： O(N 2 )
辅助空间： O(1)