806. Number of Lines To Write String

Question 806

https://leetcode.com/problems/number-of-lines-to-write-string/description/

We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an array widths, an array where widths[0] is the width of 'a', widths[1] is the width of 'b', ..., and widths[25] is the width of 'z'.

Now answer two questions: how many lines have at least one character from S, and what is the width used by the last such line? Return your answer as an integer list of length 2.

Example :
Input: 
widths = [10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
S = "abcdefghijklmnopqrstuvwxyz"
Output: [3, 60]
Explanation: 
All letters have the same length of 10. To write all 26 letters,
we need two full lines and one line with 60 units.

Example :
Input: 
widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
S = "bbbcccdddaaa"
Output: [2, 4]
Explanation: 
All letters except 'a' have the same length of 10, and 
"bbbcccdddaa" will cover 9 * 10 + 2 * 4 = 98 units.
For the last 'a', it is written on the second line because
there is only 2 units left in the first line.
So the answer is 2 lines, plus 4 units in the second line.

Answer

v1. use ord() + two conditional (26.02%)

1. 用字母的 ascii code 相減來帶出 widths

2. 如果剛好是 100 就換行

3. 如果會超過 100 也換行，並將該字母移至下行

v2. use ord() + one conditional (100%)

1. 將 'a' 的 ascii code 提出來當變數

2. 在 for 裡面只跑一個 if 當超過 100 就換行，剛好是 100 時就會在下一次相加時進 if

3. 最後再檢查一次，最後一行是不是滿了

v1

class Solution(object):
    def numberOfLines(self, widths, S):

        total,raw = 0, 1
        for s in S:
            total += widths[ord(s)-ord('a')]
            if(total == 100):
                total = 0
                raw += 1
            elif(total > 100):
                total = widths[ord(s)-ord('a')]
                raw +=1
        return [raw, total]
        

v2