refactor 554

fishercoder1534 · fishercoder1534 · commit d838d7fb2e46 · 2020-07-16T17:19:26.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_554.java b/src/main/java/com/fishercoder/solutions/_554.java
@@ -4,47 +4,18 @@
 import java.util.List;
 import java.util.Map;
 
-/**
- * 554. Brick Wall
- *
- * There is a brick wall in front of you. The wall is rectangular and has several rows of bricks.
- * The bricks have the same height but different width.
- * You want to draw a vertical line from the top to the bottom and cross the least bricks.
- * The brick wall is represented by a list of rows.
- * Each row is a list of integers representing the width of each brick in this row from left to right.
- * If your line go through the edge of a brick,
- * then the brick is not considered as crossed.
- * You need to find out how to draw the line to cross the least bricks and return the number of crossed bricks.
- * You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.
-
- Example:
- Input:
- [[1,2,2,1],
- [3,1,2],
- [1,3,2],
- [2,4],
- [3,1,2],
- [1,3,1,1]]
-
- Output: 2
- Explanation:
-
- Note:
- The width sum of bricks in different rows are the same and won't exceed INT_MAX.
- The number of bricks in each row is in range [1,10,000]. The height of wall is in range [1,10,000]. Total number of bricks of the wall won't exceed 20,000.
- */
 public class _554 {
     public static class Solution1 {
         /**
          * credit: https://leetcode.com/articles/brick-wall/
-         *
+         * <p>
          * we make use of a HashMap
          * map which is used to store entries in the form:
          * (sum,count). Here,
          * sum refers to the cumulative sum of the bricks' widths encountered in the current row, and
          * count refers to the number of times the corresponding sum is obtained. Thus,
          * sum in a way, represents the positions of the bricks's boundaries relative to the leftmost boundary.
-         *
+         * <p>
          * This is done based on the following observation:
          * We will never obtain the same value of sum twice while traversing over a particular row.
          * Thus, if the sum value is repeated while traversing over the rows, it means some row's brick boundary coincides with some previous row's brick boundary.
