806. Number of Lines To Write String
题目
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]
.
解决
通过遍历字符串S
来判断,若字母对应的宽度能够满足剩下的位置,就直接放进该行中,若不行,则新开一行来放置这些字母。
【n为字符串S
的长度】
- 时间复杂度:O(n)
- 空间复杂度:O(1)
// Runtime: 3ms
class Solution {
public:
vector<int> numberOfLines(vector<int>& widths, string S) {
vector<int> result;
int lines = 1;
int units = 0;
int len = S.length();
for (int i = 0; i < len; i++) {
if (units + widths[S[i] - 'a'] <= 100) {
// if it can fit in the rest of the positions
units += widths[S[i] - 'a'];
} else {
// if it can't fit in the rest of the positions
// creat a new lines to put it in
lines++;
units = widths[S[i] - 'a'];
}
}
result.push_back(lines);
result.push_back(units);
return result;
}
};