leetcode-806-Number of Lines To Write String

题目描述：

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].

 

要完成的函数：

vector<int> numberOfLines(vector<int>& widths, string S) 

 

说明：

1、这道题给定一个由字母组成的字符串和一个vector，vector长度为26，写着26个英文字母的宽度。要求把string中的字母写出来，按照vector中的长度写出来。

每一行最多只能有100个宽度，如果最后装不下了，那么就要写到第二行。

比如已经有了98个宽度了，现在要再写一个'a'，‘a’的宽度为4，那么这时就必须写到第二行，第一行剩下两个宽度，第二行拥有四个宽度。

要求写出string中的所有字母，返回一共有多少行，和最后一行的宽度。

2、题意清晰，这道题也就变得异常容易了。

代码如下：

    vector<int> numberOfLines(vector<int>& widths, string S) 
    {
        int s1=S.size(),sum=0,row=0;//sum记录每一行的宽度，row记录行数
        for(int i=0;i<s1;i++)
        {
            sum+=widths[S[i]-'a'];
            if(sum>100)
            {
                sum=widths[S[i]-'a'];//初始化下一行的宽度
                row++;
            }
        }
        return {row+1,sum};
    }

实测2ms，beats 100% of cpp submissions。