題目：

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.

Note:

The length of S will be in the range [1, 1000]. S will only contain lowercase letters. widths is an array of length 26. widths[i] will be in the range of [2, 10].

解釋： 26個字母，每個字母都有自己的長度，現在每一行最多隻能放下長度為100的長度，當一個字母的長度大於剩下的空位的時候，需要另起一行，求一個字串S需要佔用的總行數和最後一行需要佔用的寬度。

python程式碼：

class Solution:
    def numberOfLines(self, widths, S):
        """
        :type widths: List[int]
        :type S: str
        :rtype: List[int]
        """
        cur_width=0
        cur_line=1
        for letter in S:
            width=widths[ord(letter)-ord('a')]
            cur_line=cur_line+1 if cur_width+width>100 else cur_line
            cur_width=width if cur_width+width>100 else cur_width+width
        return cur_line,cur_width
            
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        """
        result=[]
        for i in range(left,right+1):
            if (self.isselfdividing(i)):
                result.append(i)
        return result
 

c++程式碼：

#include <vector>
class Solution {
public:
    vector<int> numberOfLines(vector<int>& widths, string S) {
        int curwidth=0;
        int curline=1;
        for (auto letter:S)
        {
            int width=widths[letter-'a'];
            curline=(curwidth+width>100?curline+1:curline);
            curwidth=(curwidth+width>100?width:curwidth+width);
            
            
        }
        vector<int> result;
        result.push_back(curline);
        result.push_back(curwidth);
        return result;
    }
};

總結： 注意要先計算curline再計算curwidth，學會使用cpp的問號表示式 還有 for (auto letter:S)。