NumberOfLinesToWriteString [source code]

public class NumberOfLinesToWriteString {
static
/******************************************************************************/
class Solution {
    public int[] numberOfLines(int[] widths, String S) {
        int line_sum = 0, line_idx = 0;
        for (int i = 0; i < S.length (); i++) {
            int cur_width = widths[S.charAt (i) - 'a'];
            if (line_sum + cur_width > 100) {
                line_sum = 0;
                line_idx++;
            }
            line_sum += cur_width;
        }
        return new int[] {line_idx + 1, line_sum};
    }
}
/******************************************************************************/

    public static void main(String[] args) {
        NumberOfLinesToWriteString.Solution tester = new NumberOfLinesToWriteString.Solution();
        int[][] inputs = {
            {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}, {3, 60},
            {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}, {2,4},
        };
        String[] string_inputs = {
            "abcdefghijklmnopqrstuvwxyz",
            "bbbcccdddaaa",
        };
        for (int i = 0; i < inputs.length / 2; i++) {
            System.out.println (Printer.separator ());
            String S = string_inputs[i];
            int[] widths = inputs[2 * i], ans = inputs[2 * i + 1], output = tester.numberOfLines (widths, S);
            String input_str = Printer.array (widths), ans_str = Printer.array (ans), output_str = Printer.array (output);
            System.out.printf ("%s and %s\n%s, expected: %s\n", 
                input_str, S, Printer.wrap (output_str, output_str.equals (ans_str) ? 92 : 91), ans_str
            );
        }
    }
}

这题的题目意思有点看不太懂;

注意我为什么这里维护index而不是line_count: 1based的数值维护起来远不如0based的简单; 所以就跟你算logprob这种东西一样, 内部的时候还是用0based的更好, 输出的时候或许再去转换就行了;

UNFINISHED

uwi:

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

Problem 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 :

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

Difficulty:Easy
Total Accepted:6.1K
Total Submissions:9.4K
Contributor:awice