refactor 806

fishercoder1534 · fishercoder1534 · commit 718928593c08 · 2020-07-23T10:36:28.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_806.java b/src/main/java/com/fishercoder/solutions/_806.java
@@ -1,62 +1,22 @@
 package com.fishercoder.solutions;
 
-/**
- * 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.
-
- 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].
-
- */
 public class _806 {
-  public static class Solution1 {
-    public int[] numberOfLines(int[] widths, String S) {
-      int numOfLines = 1;
-      int offsetInCurrentLine = 0;
-      for (char c : S.toCharArray()) {
-        if (offsetInCurrentLine + widths[c - 'a'] < 100) {
-          offsetInCurrentLine += widths[c - 'a'];
-        } else if (offsetInCurrentLine + widths[c - 'a'] == 100) {
-          numOfLines++;
-          offsetInCurrentLine = 0;
-        } else {
-          numOfLines++;
-          offsetInCurrentLine = widths[c - 'a'];
+    public static class Solution1 {
+        public int[] numberOfLines(int[] widths, String S) {
+            int numOfLines = 1;
+            int offsetInCurrentLine = 0;
+            for (char c : S.toCharArray()) {
+                if (offsetInCurrentLine + widths[c - 'a'] < 100) {
+                    offsetInCurrentLine += widths[c - 'a'];
+                } else if (offsetInCurrentLine + widths[c - 'a'] == 100) {
+                    numOfLines++;
+                    offsetInCurrentLine = 0;
+                } else {
+                    numOfLines++;
+                    offsetInCurrentLine = widths[c - 'a'];
+                }
+            }
+            return new int[]{numOfLines, offsetInCurrentLine};
         }
-      }
-      return new int[] {numOfLines, offsetInCurrentLine};
     }
-  }
 }
