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].

這道題給了我們一個字串，讓我們把裡面的字母寫下來，規定了每一行的長度為100，然後每個字母的長度可以在widths陣列中查詢，說是如果某一個字母加上後超過了長度100的限制，那麼就移動到下一行，問我們最終需要多少行，和最後一行的長度。這道題並沒有太大的難度和技巧，就是楞頭寫唄，遍歷所有的字母，然後查表得到其寬度，然後看加上這個新寬度是否超了100，超了的話，行數計數器自增1，並且當前長度為這個字母的長度，因為另起了一行。如果沒超100，那麼行長度就直接加上這個字母的長度。遍歷完成後返回行數和當前行長度即可，參見程式碼如下：

class Solution {
public:
    vector<int> numberOfLines(vector<int>& widths, string S) {
        int cnt = 1, cur = 0;
        for (char c : S) {
            int t = widths[c - 'a'];
            if (cur + t > 100) ++cnt;
            cur = (cur + t > 100) ? t : cur + t;
        }
        return {cnt, cur};
    }
};

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