Leetcode 806 Number of Lines To Write String 字符串的行数

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].

翻译：

    1.给出一个长度26的数组widths，保存着26个字母打印出来所需要的的屏幕宽度，而每行屏幕宽度都只有100，超过这个宽度的话，字母只能打印到下一行去，给出一段字符串，求出打印后最后一个字母所在的行数，列数。

思路：

    1.打印字符串，记录行数lines，列数pos，如果pos超过100了，打印到下一行去

    2.时间复杂度O（n）

代码：

class Solution {
public:
    vector<int> numberOfLines(vector<int>& widths, string S) 
    {
        int n=S.size(),lines=1,pos=0;
        for(int i=0;i<n;i++)
        {
            pos += widths[S[i]-'a'];
            if(pos>100) //写出头了
            {
                lines++;pos=0;//往下一行写
                i--;//就当没写过这个单词
            }
        }
        return {lines,pos};
    }
};