给定一个N * M网格。它正好包含三个“*” ，其他每个单元格都是一个“.”。其中“*”表示矩形的顶点。任务是找到第四个（缺失的）顶点的坐标（基于 1 的索引）。
例子：
Input: grid[][] = {“*.*”, “*..”, “…”}
Output: 2 3
(1, 1), (1, 3) and (2, 1) are the given coordinates of the rectangle where (2, 3) is the missing coordinate.
Input: grid[][] = {“*.*”, “..*”}
Output: 2 1
方法：通过迭代给定的网格来找到 3 个顶点的坐标。由于矩形由 2 个 X 坐标和 2 个 Y 坐标以及 4 个顶点组成，因此每个 X 坐标和 Y 坐标都应该恰好出现两次。我们可以计算每个 X 和 Y 坐标在 3 个给定顶点中出现的次数，而第 4 个将具有仅出现一次的坐标。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that return the coordinates
// of the fourth vertex of the rectangle
pair findFourthVertex(int n, int m, string s[])
{
map row, col;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
// Save the coordinates of the given
// vertices of the rectangle
if (s[i][j] == '*') {
row[i]++;
col[j]++;
}
int x, y;
for (auto tm : row)
if (tm.second == 1)
x = tm.first;
for (auto tm : col)
if (tm.second == 1)
y = tm.first;
// 1-based indexing
return make_pair(x + 1, y + 1);
}
// Driver code
int main()
{
string s[] = { "*.*", "*..", "..." };
int n = sizeof(s) / sizeof(s[0]);
int m = s[0].length();
auto rs = findFourthVertex(n, m, s);
cout << rs.first << " " << rs.second;
} Java
// Java implementation of the approach
import java.util.HashMap;
import java.util.Map;
class GfG
{
// Function that return the coordinates
// of the fourth vertex of the rectangle
static Pair findFourthVertex(int n,
int m, String s[])
{
HashMap row = new HashMap<>();
HashMap col = new HashMap<>();
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Save the coordinates of the given
// vertices of the rectangle
if (s[i].charAt(j) == '*')
{
if (row.containsKey(i))
{
row.put(i, row.get(i) + 1);
}
else
{
row.put(i, 1);
}
if (col.containsKey(j))
{
col.put(j, col.get(j) + 1);
}
else
{
col.put(j, 1);
}
}
}
}
int x = 0, y = 0;
for (Map.Entry entry : row.entrySet())
{
if (entry.getValue() == 1)
x = entry.getKey();
}
for (Map.Entry entry : col.entrySet())
{
if (entry.getValue() == 1)
y = entry.getKey();
}
// 1-based indexing
Pair ans = new Pair<>(x + 1, y + 1);
return ans;
}
// Driver Code
public static void main(String []args)
{
String s[] = { "*.*", "*..", "..." };
int n = s.length;
int m = s[0].length();
Pair rs = findFourthVertex(n, m, s);
System.out.println(rs.first + " " + rs.second);
}
}
class Pair
{
A first;
B second;
public Pair(A first, B second)
{
this.first = first;
this.second = second;
}
}
// This code is contributed by Rituraj Jain Python3
# Python3 implementation of the approach
# Function that return the coordinates
# of the fourth vertex of the rectangle
def findFourthVertex(n, m, s) :
row = dict.fromkeys(range(n), 0)
col = dict.fromkeys(range(m), 0)
for i in range(n) :
for j in range(m) :
# Save the coordinates of the given
# vertices of the rectangle
if (s[i][j] == '*') :
row[i] += 1;
col[j] += 1;
for keys,values in row.items() :
if (values == 1) :
x = keys;
for keys,values in col.items() :
if (values == 1) :
y = keys;
# 1-based indexing
return (x + 1, y + 1) ;
# Driver code
if __name__ == "__main__" :
s = [ "*.*", "*..", "..." ]
n = len(s);
m = len(s[0]);
rs = findFourthVertex(n, m, s);
print(rs[0], rs[1])
# This code is contributed by RyugaC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GfG
{
// Function that return the coordinates
// of the fourth vertex of the rectangle
static Pair findFourthVertex(int n,
int m, String []s)
{
Dictionary row = new Dictionary();
Dictionary col = new Dictionary();
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Save the coordinates of the given
// vertices of the rectangle
if (s[i][j] == '*')
{
if (row.ContainsKey(i))
{
row[i] = row[i] + 1;
}
else
{
row.Add(i, 1);
}
if (col.ContainsKey(j))
{
col[j] = col[j] + 1;
}
else
{
col.Add(j, 1);
}
}
}
}
int x = 0, y = 0;
foreach(KeyValuePair entry in row)
{
if (entry.Value == 1)
x = entry.Key;
}
foreach(KeyValuePair entry in col)
{
if (entry.Value == 1)
y = entry.Key;
}
// 1-based indexing
Pair ans = new Pair(x + 1, y + 1);
return ans;
}
// Driver Code
public static void Main(String []args)
{
String []s = { "*.*", "*..", "..." };
int n = s.Length;
int m = s[0].Length;
Pair rs = findFourthVertex(n, m, s);
Console.WriteLine(rs.first + " " + rs.second);
}
}
class Pair
{
public A first;
public B second;
public Pair(A first, B second)
{
this.first = first;
this.second = second;
}
}
// This code is contributed by 29AjayKumar Javascript
输出：
2 3时间复杂度： O(N * M)
辅助空间： O(N + M)
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