QUESTION:
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]
.
EXPLANATION:
先写出逻辑吧:
1.首先每个字母代表的长度不一样,那得到每个字母的长度 widths[chars[i]=’a’]
2.然后累加,大于100的重新换行
3.换行就是行数加一,并且新行的unit已经是一个字符的长度了。
根据逻辑写出来即可。
SOLUTION:
class Solution {
public int[] numberOfLines(int[] widths, String S) {
char[] chars = S.toCharArray();
int line = 1;
int count = 0;
for(int i = 0;i<chars.length;i++){
int len = widths[chars[i]-'a'];
count+=len;
if(count>100){
line++;
count=len;
}
}
return new int[]{line,count};
}
}