806. Number of Lines To Write String

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.

题目解释：这道题意思是按照a对应widths里面index为0的长度，b对应widths里面index为1的长度的规则，将S中每一个字符的长度相加，超过100时连同当前字符整个宽度进到下一行，最后求行数和最后行的宽度

//Time: O(n), Space: O(1)   
 public int[] numberOfLines(int[] widths, String S) {
        if (widths == null || S == null || widths.length == 0 || S.length() == 0) {
            return null;
        }
    
        int count = 1;//注意行数要从1开始
        int length = 0;
        
        for (char c : S.toCharArray()) {
            length += widths[c - 'a'];
            
            if (length > 100) {
                count++;
                length = widths[c - 'a'];//当宽度大于100时要reset新一行的宽度
            } 
        }
        
        return new int[]{count, length};
    }