refactor 1377

fishercoder1534 · fishercoder1534 · commit f9743255a66a · 2020-11-06T06:32:34.000-08:00
diff --git a/src/main/java/com/fishercoder/solutions/_1377.java b/src/main/java/com/fishercoder/solutions/_1377.java
@@ -7,47 +7,11 @@
 import java.util.Map;
 import java.util.Queue;
 
-/**
- * 1377. Frog Position After T Seconds
- *
- * Given an undirected tree consisting of n vertices numbered from 1 to n.
- * A frog starts jumping from the vertex 1.
- * In one second, the frog jumps from its current vertex to another unvisited vertex if they are directly connected.
- * The frog can not jump back to a visited vertex.
- * In case the frog can jump to several vertices it jumps randomly to one of them with the same probability,
- * otherwise, when the frog can not jump to any unvisited vertex it jumps forever on the same vertex.
- * The edges of the undirected tree are given in the array edges, where edges[i] = [fromi, toi] means that exists an edge connecting directly the vertices fromi and toi.
- * Return the probability that after t seconds the frog is on the vertex target.
- *
- * Example 1:
- * Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
- * Output: 0.16666666666666666
- * Explanation: The figure above shows the given graph.
- * The frog starts at vertex 1, jumping with 1/3 probability to the vertex 2 after second 1 and
- * then jumping with 1/2 probability to vertex 4 after second 2.
- * Thus the probability for the frog is on the vertex 4 after 2 seconds is 1/3 * 1/2 = 1/6 = 0.16666666666666666.
- *
- * Example 2:
- * Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7
- * Output: 0.3333333333333333
- * Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 = 0.3333333333333333 probability to the vertex 7 after second 1.
- *
- * Example 3:
- * Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 20, target = 6
- * Output: 0.16666666666666666
- *
- * Constraints:
- * 1 <= n <= 100
- * edges.length == n-1
- * edges[i].length == 2
- * 1 <= edges[i][0], edges[i][1] <= n
- * 1 <= t <= 50
- * 1 <= target <= n
- * Answers within 10^-5 of the actual value will be accepted as correct.
- * */
 public class _1377 {
     public static class Solution1 {
-        /**credit: https://leetcode.com/problems/frog-position-after-t-seconds/discuss/532505/Java-Straightforward-BFS-Clean-code-O(N)*/
+        /**
+         * credit: https://leetcode.com/problems/frog-position-after-t-seconds/discuss/532505/Java-Straightforward-BFS-Clean-code-O(N)
+         */
         public double frogPosition(int n, int[][] edges, int t, int target) {
             List<Integer>[] graph = new ArrayList[n];
             for (int i = 0; i < n; i++) {
