字符串中的垂直顺序遍历

给定一个由空格分隔的多个单词组成的字符串str ，任务是打印该字符串的垂直顺序遍历。垂直顺序遍历是指当每个单词按垂直顺序插入网格时，对给定字符串形成的网格的行遍历。
注意：空格(“”)由“ $ ”符号表示，以便更好地可视化。

例子：

Input: str = “geeks for geeks”
Output: [gfg, eoe, ere, k$k, s$s]
Explanation: Vertical order of the string will result in the following column values.
g f g
e o e
e r e
k $ k
s $ s
Now the Answer will be a list of strings appearing row-wise.

Input: str = “ge ek sfor ge eks”
Output: [gesge, ekfek, $$o$s, $$r$$]

方法：想法是使用HashMap，其中hashmap的键代表索引，值代表该索引处的字符。可以按照以下步骤计算垂直顺序遍历。

1. 将给定的字符串拆分为单词。
   例如：

   If String is "Geeks For Geeks"
   then the words will be
   words = {"Geeks", "For", "Geeks"}
   

2. 最初，找到最大长度的单词。然后所有的单词都小于或等于这个长度。
3. 然后创建一个最大单词大小的HashMap。
   例如：

   If the maximum length word is 5,
   then the HashMap will be
   map = {
           {0, ""},
           {1, ""},
           {2, ""},
           {3, ""},
           {4, ""}
         }
   

4. 对于每个单词，然后检查该单词是否具有此最大长度。如果是，则将其附加到 HashMap。否则，额外的空格(表示为“ $ ”)被添加到字符串。
   例如：

   For String = "Geeks For Geeks"
   then the HashMap after word "Geeks"
   map = {
           {0, "G"},
           {1, "e"},
           {2, "e"},
           {3, "k"},
           {4, "s"}
         }
   
   then the HashMap after word "For"
   map = {
           {0, "GF"},
           {1, "eo"},
           {2, "er"},
           {3, "k$"},
           {4, "s$"}
         }
   
   then the HashMap after word "Geeks"
   map = {
           {0, "GFG"},
           {1, "eoe"},
           {2, "ere"},
           {3, "k$k"},
           {4, "s$s"}
         }
   

5. HashMap 中每个索引处的单词表示给定字符串的最终垂直顺序遍历中的单词。
6. 因此，最后，从 HashMap 中获取(从每个索引)字符串并添加到数组中，然后返回数组。
   例如：

   Vertial order traversal of the above example:
   "GFG eoe ere k$k s$s"
   

下面是上述方法的实现：

Java

// Java implementation of the above approach.
  
import java.util.*;
  
class GFG {
  
    // Function to return the vertical order
    // traversal of String
    public static String printVertically(String s)
    {
        String result = "";
  
        // Hashmap to store the indices
        // of the characters
        HashMap map
            = new HashMap<>();
  
        Scanner sc = new Scanner(s);
        int max = Integer.MIN_VALUE;
  
        // Finding maximum length word
        // from given string
        while (sc.hasNext()) {
            String st = sc.next();
            max = Math.max(max, st.length());
        }
  
        sc = new Scanner(s);
        while (sc.hasNext()) {
            String st = sc.next();
  
            // Inserting strings in the hashmap
            // in vertical fashion
            // based on character index
            for (int i = 0; i < st.length(); i++) {
                char ch = st.charAt(i);
                map.put(i, map.getOrDefault(i, "") + ch);
            }
  
            // If the string is smaller than the
            // maximum length string, then
            // add spaces in the string
            for (int i = st.length(); i < max; i++) {
                map.put(i,
                        map.getOrDefault(i,
                                         "")
                            + "$");
            }
        }
  
        // Adding  all String values in the list
        for (int i = 0; i < max; i++) {
            String a = map.get(i);
            result += a + " ";
        }
        return result;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String st = "ge ek sfor ge eks";
        System.out.println(printVertically(st));
    }
}

Python3

# Python3 implementation of the above approach.
  
# Function to return the vertical order
# traversal of
def printVertically(s):
    s = s.split(" ")
    result = ""
      
    # Hashmap to store the indices
    # of the characters
    map = dict()
    mx = 0
  
    # Finding maximum length word
    # from given string
    for i in s:
        mx = max(mx, len(i))
    n = mx
    i = 0
    while (i < mx):
          
        # Inserting strings in the hashmap
        # in vertical fashion
        # based on character index
        for j in s:
            if i >= len(j):
                map[i] = map.get(i, "") + "$"
            else:
                map[i] = map.get(i, "") + j[i]
        i += 1    
  
    # Adding all values in the list
    for j in range(mx):
        result += map[j] + " "
    return result
  
# Driver code
if __name__ == '__main__':
    st = "ge ek sfor ge eks"
    print(printVertically(st))
  
# This code is contributed by mohit kumar 29

C#

// C# implementation of the above approach.
using System;
using System.Collections.Generic;
  
class GFG
{
  
    // Function to return the vertical order
    // traversal of String
    public static String printVertically(String s)
    {
        String result = "";
  
        // Hashmap to store the indices
        // of the characters
        Dictionary map
            = new Dictionary();
  
        String []sc = s.Split(' ');
        int max = int.MinValue;
        int i = 0;
          
        // Finding maximum length word
        // from given string
        while (i

---

输出：

gesge ekfek $$o$s $$r$$

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