LeetCode-Number of Lines To Write String

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_24133491/article/details/82796621

Description：
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 1:

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.

Example2 :

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 length 26.
• widths[i] will be in the range of [2, 10].

题意：给定一个字符串及其一个宽度数组，表示每个小写字母需要占据的宽度；要求计算给定的字符串可以排列多少行（每行的最大宽度为100）及最后一行的宽度；

解法：就是通过遍历字符串，当加上当前字母的宽度会超过此时允许的最大宽度时，排列到下一行；利用两个变量用于统计行数，及当前行的宽度；

Java

class Solution {
    public int[] numberOfLines(int[] widths, String S) {
        int lastLen = 0;
        int row = 1;
        for (int i = 0; i < S.length();) {
            if (lastLen + widths[S.charAt(i) - 'a'] > 100) {
                row += 1;
                lastLen = 0;
            } else {
                lastLen += widths[S.charAt(i) - 'a'];
                i++;
            }
        }
        return new int[] {row, lastLen};
    }
}

所有的代码都保存在GitHub；