从字符矩阵中删除行或列重复
给定一个仅包含小写字母的字符矩阵，从矩阵中形成一个新字符串，该字符串在删除每一行和每一列的重复字符后仍然存在。其余字符按行连接以形成字符串。
例子：
Input : Matrix of characters (Or array
of strings)
aabcd
effgh
iijkk
lmnoo
pqqrs
ttuvw
xxyyz
Output : bcdeghjlmnprsuvwz
The characters that appear more than once
in their row or column are removed.
Input : zx
xz
Output :zxxz方法：遍历整个矩阵，对于每个元素，检查其对应的行和列。如果有重复，则将另一个矩阵中的特定单元格标记为真。最后，将另一个矩阵中存在假值的所有字符连接起来形成字符串。
C++
// CPP code to form string after removing
// duplicates from rows and columns.
#include
using namespace std;
// Function to check duplicates
// in row and column
void findDuplciates(string a[], int n, int m)
{
// Create an array isPresent and initialize
// all entries of it as false. The value of
// isPresent[i][j] is going to be true if
// s[i][j] is present in its row or column.
bool isPresent[n][m];
memset(isPresent, 0, sizeof(isPresent));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Checking every row for
// duplicates of a[i][j]
for (int k = 0; k < n; k++) {
if (a[i][j] == a[k][j] && i != k) {
isPresent[i][j] = true;
isPresent[k][j] = true;
}
}
// Checking every column for
// duplicate characters
for (int k = 0; k < m; k++) {
if (a[i][j] == a[i][k] && j != k) {
isPresent[i][j] = true;
isPresent[i][k] = true;
}
}
}
}
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
// If the character is unique
// in its row and column
if (!isPresent[i][j])
printf("%c", a[i][j]);
}
// Driver code
int main()
{
int n = 2, m = 5;
// character array
string a[] = { "zx", "xz" };
// Calling function
findDuplciates(a, n, m);
return 0;
} Java
// Java code to form string
// after removing duplicates
// from rows and columns.
class GFG
{
// Function to check
// duplicates in row
// and column
static void findDuplciates(String []a,
int n, int m)
{
// Create an array isPresent
// and initialize all entries
// of it as false. The value
// of isPresent[i,j] is going
// to be true if s[i,j] is
// present in its row or column.
boolean [][]isPresent = new boolean[n] [m];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
isPresent[i][j]=false;
}
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Checking every row for
// duplicates of a[i,j]
for (int k = 0; k < n; k++)
{
if (a[i].charAt(j)== a[k].charAt(j) &&
i != k)
{
isPresent[i][j] = true;
isPresent[k][j] = true;
}
}
// Checking every column for
// duplicate characters
for (int k = 0; k < m; k++)
{
if (a[i].charAt(j)== a[i].charAt(k) &&
j != k)
{
isPresent[i][j] = true;
isPresent[i][k] = true;
}
}
}
}
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
// If the character is
// unique in its row
// and column
if (isPresent[i][j]==false)
System.out.print(a[i].charAt(j));
}
// Driver code
public static void main(String []args)
{
int n = 2, m = 2;
// character array
String []a = new String[]{"zx",
"xz"};
// Calling function
findDuplciates(a, n, m);
}
}
// This code is contributed by
// Hritik Raj( ihritik)Python3
# Python3 code to form string after removing
# duplicates from rows and columns.
# Function to check duplicates
# in row and column
def findDuplicates(a, n, m):
# Create an array isPresent and initialize
# all entries of it as false. The value of
# isPresent[i][j] is going to be true if
# s[i][j] is present in its row or column.
isPresent = [[False for i in range(n)]
for j in range(m)]
for i in range(n):
for j in range(m):
# Checking every row for
# duplicates of a[i][j]
for k in range(n):
if i != k and a[i][j] == a[k][j]:
isPresent[i][j] = True
isPresent[k][j] = True
# Checking every row for
# duplicates of a[i][j]
for k in range(m):
if j != k and a[i][j] == a[i][k]:
isPresent[i][j] = True
isPresent[i][k] = True
for i in range(n):
for j in range(m):
# If the character is unique
# in its row and column
if not isPresent[i][j]:
print(a[i][j], end = "")
# Driver Code
if __name__ == "__main__":
n = 2
m = 2
# character array
a = ["zx", "xz"]
# Calling function
findDuplicates(a, n, m)
# This code is contributed by
# sanjeev2552C#
// C# code to form string
// after removing duplicates
// from rows and columns.
using System;
class GFG
{
// Function to check
// duplicates in row
// and column
static void findDuplciates(string []a,
int n, int m)
{
// Create an array isPresent
// and initialize all entries
// of it as false. The value
// of isPresent[i,j] is going
// to be true if s[i,j] is
// present in its row or column.
bool [,]isPresent = new bool[n, m];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Checking every row for
// duplicates of a[i,j]
for (int k = 0; k < n; k++)
{
if (a[i][j] == a[k][j] &&
i != k)
{
isPresent[i, j] = true;
isPresent[k, j] = true;
}
}
// Checking every column for
// duplicate characters
for (int k = 0; k < m; k++)
{
if (a[i][j] == a[i][k] &&
j != k)
{
isPresent[i, j] = true;
isPresent[i, k] = true;
}
}
}
}
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
// If the character is
// unique in its row
// and column
if (!isPresent[i, j])
Console.Write(a[i][j]);
}
// Driver code
static void Main()
{
int n = 2, m = 2;
// character array
string []a = new string[]{"zx",
"xz"};
// Calling function
findDuplciates(a, n, m);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)Javascript
输出：
zxxz