806 Number of Lines To Write String

You are given a string s of lowercase English letters and an array widths denoting how many pixels wide each lowercase English letter is. Specifically, widths[0] is the width of 'a', widths[1] is the width of 'b', and so on.

You are trying to write s across several lines, where each line is no longer than 100 pixels. Starting at the beginning of s, write as many letters on the first line such that the total width does not exceed 100 pixels. Then, from where you stopped in s, continue writing as many letters as you can on the second line. Continue this process until you have written all of s.

Return an array result of length 2 where:

• result[0] is the total number of lines.

• result[1] is the width of the last line in pixels.

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: You can write s as follows:
abcdefghij  // 100 pixels wide
klmnopqrst  // 100 pixels wide
uvwxyz      // 60 pixels wide
There are a total of 3 lines, and the last line is 60 pixels wide.

Example 2:

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: You can write s as follows:
bbbcccdddaa  // 98 pixels wide
a            // 4 pixels wide
There are a total of 2 lines, and the last line is 4 pixels wide.

Constraints:

• widths.length == 26

• 2 <= widths[i] <= 10

• 1 <= s.length <= 1000

• s contains only lowercase English letters.

看了半天才看懂了题目，差点勾起了68 Text justification的恐惧，毕竟是狗的题。发现狗挺喜欢考这种边界很难把控的题。这里主要难点是，什么时候到下一行。如果我们发现加的长度大于100了，就要减回来。因为过一遍，所以T:O(n), S:O(1)

public int[] numberOfLines(int[] widths, String s) {
    int[] result = new int[2];
    if (widths == null || widths.length == 0 || s == null || s.isEmpty()) {
        return null;
    }

    int lastWidth = 0;
    int rowCnt = 0;

    int stringInd = 0;
    while (stringInd < s.length()) {
        lastWidth = 0;
        while (stringInd < s.length() && lastWidth < 100) {
            char curChar = s.charAt(stringInd);
            lastWidth = lastWidth + widths[curChar - 'a'];
            stringInd++;
        }

        if (lastWidth > 100) {
            stringInd--;
        }

        rowCnt++;
    }

    result[0] = rowCnt;
    result[1] = lastWidth;

    return result;
}