806. Number of Lines To Write String（python+cpp）

题目：

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].

解释：
26个字母，每个字母都有自己的长度，现在每一行最多只能放下长度为100的长度，当一个字母的长度大于剩下的空位的时候，需要另起一行，求一个字符串S需要占用的总行数和最后一行需要占用的宽度。

python代码：

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++代码：

#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;
    }
};

总结：
注意要先计算curline再计算curwidth，学会使用cpp的问号表达式还有 for (auto letter:S)。

806. Number of Lines To Write String（python+cpp）

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