写入给定String所需的行数

给定一个字符串str和一个整数数组width [] ，其中：

width[0] = width of character ‘a’
width[1] = width of character ‘b’
…
width[25] = width of character ‘z’

我们的任务是找到将字符串str写在纸上所需的行数，以及占据该行的最后一行的宽度。
注意：线的宽度是10个单位。

例子：

Input: str = “bbbcccdddaa”,
width[] = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
Output: (2, 8)
“bbbcccddd” will cover first line (9 * 1 = 9 units)
As ‘a’ has a width of 4 which cannot fit the remaining 1 unit in the first line.
It’ll have to be written in the second line.
So, next line will contain “aa” covering 4 * 2 = 8 units.
We need 1 full line and one line with width 8 units.

Input: str = “abcdefghijklmnopqrstuvwxyz”,
width[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
Output: (3, 6)
All the characters have the same width of 1. To write all 26 characters,
We need 2 full lines and one line with width 6 units.

方法：我们将字符串str中的每个字符一一写入。当我们编写一个字符，我们会立即更新(线，宽)，以跟踪到现在为止我们已经使用了多少行，以及最后一行中已用空间的长度是多少。
如果str中的width [char]符合我们当前的行，我们将添加它。否则，我们将从新行开始

下面是上述方法的实现：

C++

// CPP implementation of the approach
#include 
using namespace std;
  
// Function to return the number of lines required
pair numberOfLines(string S, int *widths)
{
    // If string is empty
    if (S.empty())
        return {0, 0};
  
    // Initialize lines and width
    int lines = 1, width = 0;
  
    // Iterate through S
    for (auto character : S)
    {
        int w = widths[character - 'a'];
        width += w;
  
        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }
  
    // Return lines and width used
    return {lines, width};
}
  
// Driver Code
int main()
{
    string S = "bbbcccdddaa";
    int widths[] = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
  
    // Function call to print required answer
    pair ans = numberOfLines(S, widths);
    cout << ans.first << " " << ans.second << endl;
  
    return 0;
}
  
// This code is contributed by
// sanjeev2552

Java

// JAVA implementation of the approach
class GFG
{
  
// Function to return the number of lines required
static int[] numberOfLines(String S, int []widths)
{
    // If String is empty
    if (S.isEmpty())
        return new int[]{0, 0};
  
    // Initialize lines and width
    int lines = 1, width = 0;
  
    // Iterate through S
    for (char character : S.toCharArray())
    {
        int w = widths[character - 'a'];
        width += w;
  
        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }
  
    // Return lines and width used
    return new int[]{lines, width};
}
  
// Driver Code
public static void main(String[] args)
{
    String S = "bbbcccdddaa";
    int widths[] = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
  
    // Function call to print required answer
    int []ans = numberOfLines(S, widths);
    System.out.print(ans[0]+ " " + ans[1] +"\n");
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
  
# Function to return the number of lines required
def numberOfLines(S, widths):
  
    # If string is empty
    if(S == ""):
        return 0, 0
  
    # Initialize lines and width
    lines, width = 1, 0
  
    # Iterate through S
    for c in S:
        w = widths[ord(c) - ord('a')]
        width += w
        if width > 10:
            lines += 1
            width = w
  
    # Return lines and width used
    return lines, width
  
  
# Driver Code
S = "bbbcccdddaa"
Widths = [4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
  
# Function call to print required answer
print(numberOfLines(S, Widths))

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the number of lines required
static int[] numberOfLines(String S, int []widths)
{
    // If String is empty
    if (S.Length == 0)
        return new int[]{0, 0};
  
    // Initialize lines and width
    int lines = 1, width = 0;
  
    // Iterate through S
    foreach (char character in S.ToCharArray())
    {
        int w = widths[character - 'a'];
        width += w;
  
        if (width >= 10)
        {
            lines++;
            width = w;
        }
    }
  
    // Return lines and width used
    return new int[]{lines, width};
}
  
// Driver Code
public static void Main(String[] args)
{
    String S = "bbbcccdddaa";
    int []widths = {4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
  
    // Function call to print required answer
    int []ans = numberOfLines(S, widths);
    Console.Write(ans[0]+ " " + ans[1] +"\n");
}
}
  
// This code is contributed by 29AjayKumar

输出：

(2, 8)