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 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 = 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};
}
};
参考资料:
https://leetcode.com/problems/number-of-lines-to-write-string/solution/
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