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
Swill be in the range [1, 1000]. -
Swill only contain lowercase letters. -
widthsis an array of length26. -
widths[i]will be in the range of[2, 10].
这道题给了我们一个字符串，让我们把里面的字母写下来，规定了每一行的长度为100，然后每个字母的长度可以在widths数组中查询，说是如果某一个字母加上后超过了长度100的限制，那么就移动到下一行，问我们最终需要多少行，和最后一行的长度。这道题并没有太大的难度和技巧，就是楞头写呗，遍历所有的字母，然后查表得到其宽度，然后看加上这个新宽度是否超了100，超了的话，行数计数器自增1，并且当前长度为这个字母的长度，因为另起了一行。如果没超100，那么行长度就直接加上这个字母的长度。遍历完成后返回行数和当前行长度即可，参见代码如下：
class Solution {
public:
vector<int> numberOfLines(vector<int>& widths, string S) {
int cnt = , cur = ;
for (char c : S) {
int t = widths[c - 'a'];
if (cur + t > ) ++cnt;
cur = (cur + t > ) ? t : cur + t;
}
return {cnt, cur};
}
};
参考资料：
https://leetcode.com/problems/number-of-lines-to-write-string/solution/