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]
.
字符串的占位统计。
每行有100个空位,widths表示每个字符所占的空间大小,如果末尾不能容纳一个字符,那么就将其移动至下一行。
统计字符串占据的行数及最后一行所占的空间大小。
class Solution { public: vector<int> numberOfLines(vector<int>& widths, string S) { int m = 0, n = 0; for (int i = 0; i < S.size(); ++i) { m += widths[(S[i]-‘a‘)]; if (m >= 100) { if (m == 100) { n++; m = 0; } else { n++; m = widths[(S[i]-‘a‘)]; } } } return {n+1, m}; } };
原文地址:https://www.cnblogs.com/immjc/p/9149051.html
时间: 2024-10-08 15:38:40