leetCode806. The number of rows required to write a string

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We want to write the given string S from left to right on each line, the maximum width of each row is 100 units, if we write a letter that will make this line more than 100 units, then we should write this letter to the next line. We given an array of widths, this array widths[0] stands for ' A ' required units, Widths[1] stands for ' B ' required units, ..., widths[25] represents the units required for ' Z '.

Now answer two questions: at least how many lines can put down s, and how many units the last line is used. Return your answer as a list of integers 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]
explanation: 
All characters have the same occupancy unit 10. So to write all 26 letters,
we need 2 full lines and a row of 60 units.

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: 
Removing all characters of the letter ' A ' is the same unit 10, and the string" Bbbcccdddaa "will overwrite 9 * 10 + 2 * 4 = 98 units.
The last letter ' A ' will be written to the second line because there are only 2 units left in the first line.
So, the answer is 2 lines, and the second line has 4 units of width.

Note: The length of the string S is in the range of [1, 1000]. S contains only lowercase letters. Widths is an array of length 26. Solution Idea: Define 3 variables, num represents the total units per row, line represents the number of rows, r represents each occupancy unit, and then uses for to connect each string to the corresponding unit.

Import Java.util.Arrays;
		public class NumberOfLines {public static int[] NumberOfLines (int[] widths, String S) {int num = 0;
		int line=1;
			for (int i=0;i<s.length (); i++) {int r = 0;
				Switch (S.charat (i)) {case ' a ': num+=widths[0];
				R+=widths[0];
			Break
				Case ' B ': num+=widths[1];
				R+=WIDTHS[1];
			Break
				Case ' C ': num+=widths[2];
				R+=WIDTHS[2];
			Break
				Case ' d ': num+=widths[3];
				R+=WIDTHS[3];
			Break
				Case ' E ': num+=widths[4];
				R+=WIDTHS[4];
			Break
				Case ' F ': num+=widths[5];
				R+=WIDTHS[5];
			Break
				Case ' G ': num+=widths[6];
				R+=WIDTHS[6];
			Break
				Case ' H ': num+=widths[7];
				R+=WIDTHS[7];
			Break
				Case ' I ': num+=widths[8];
				R+=WIDTHS[8];
			Break
				Case ' J ': num+=widths[9];
				R+=WIDTHS[9];
			Break
				Case ' K ': num+=widths[10];
				R+=WIDTHS[10];
			Break
				Case ' l ': num+=widths[11];
				R+=WIDTHS[11];
			Break Case ' m ': num+=widths[12];
				R+=WIDTHS[12];
			Break
				Case ' n ': num+=widths[13];
				R+=WIDTHS[13];
			Break
				Case ' O ': num+=widths[14];
				R+=WIDTHS[14];
			Break
				Case ' P ': num+=widths[15];
				R+=WIDTHS[15];
			Break
				Case ' Q ': num+=widths[16];
				R+=WIDTHS[16];
			Break
				Case ' R ': num+=widths[17];
				R+=WIDTHS[17];
			Break
				Case ' s ': num+=widths[18];
				R+=WIDTHS[18];
			Break
				Case ' t ': num+=widths[19];
				R+=WIDTHS[19];
			Break
				Case ' u ': num+=widths[20];
				R+=WIDTHS[20];
			Break
				Case ' V ': num+=widths[21];
				R+=WIDTHS[21];
			Break
				Case ' W ': num+=widths[22];
				R+=WIDTHS[22];
			Break
				Case ' x ': num+=widths[23];
				R+=WIDTHS[23];
			Break
				Case ' Y ': num+=widths[24];
				R+=WIDTHS[24];
			Break
				Case ' Z ': num+=widths[25];
				R+=WIDTHS[25];
			Break
			Default:break;
				} if (num>100) {num=0;
				line++;
			Num+=r;
		}} int[] a = {line,num};
	return A; } PubLic static void Main (string[] args) {int[] 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};
		String S = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
	System.out.println (arrays.tostring (NumberOfLines (widths, S)); }