LeetCode Frog Jump

原題連結在這裡：https://leetcode.com/problems/frog-jump/description/

題目：

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.

Given a list of stones' positions (in units) in sorted ascending order, determine if the frog is able to cross the river by landing on the last stone. Initially, the frog is on the first stone and assume the first jump must be 1 unit.

If the frog's last jump was k units, then its next jump must be either k - 1, k, or k + 1 units. Note that the frog can only jump in the forward direction.

Note:

• The number of stones is ≥ 2 and is < 1,100.
• Each stone's position will be a non-negative integer < 231.
• The first stone's position is always 0.

Example 1:

[0,1,3,5,6,8,12,17]

There are a total of 8 stones.
The first stone at the 0th unit, second stone at the 1st unit,
third stone at the 3rd unit, and so on...
The last stone at the 17th unit.

Return true. The frog can jump to the last stone by jumping 
1 unit to the 2nd stone, then 2 units to the 3rd stone, then 
2 units to the 4th stone, then 3 units to the 6th stone, 
4 units to the 7th stone, and 5 units to the 8th stone.      

Example 2:

[0,1,2,3,4,8,9,11]

Return false. There is no way to jump to the last stone as 
the gap between the 5th and 6th stone is too large.      

題解：

看青蛙是否能正好跳到最後一塊石頭上.

DP問題，需要保留的曆史資訊是在目前這塊石頭上能向前跳多少步. 

狀态轉移時，若在目前石頭向前跳n步時剛好能落到後面的石頭上，那麼後面石頭就能能跳n-1, n , n+1步，把這三種情況加到後面那塊石頭的存儲資訊裡.

答案是看中間有沒有正好跳到最後一塊石頭上的可能.

Time Complexity: O(n^2). n = stones.length. 最壞情況每個step都能跳到，那麼對于後面的石頭就有接近3n種可向前跳的步數.

1 class Solution {
 2     public boolean canCross(int[] stones) {
 3         if(stones == null || stones.length == 0){
 4             return true;
 5         }
 6         
 7         int len = stones.length;
 8         HashMap<Integer, HashSet<Integer>> hm = new HashMap<Integer, HashSet<Integer>>();
 9         for(int i = 0; i<len; i++){
10             hm.put(stones[i], new HashSet<Integer>());
11         }
12         
13         hm.get(stones[0]).add(1);
14         for(int i = 0; i<len; i++){
15             HashSet<Integer> steps = hm.get(stones[i]);
16             for(int step : steps){
17                 if(stones[i]+step == stones[len-1]){
18                     return true;
19                 }
20                 if(hm.containsKey(stones[i]+step)){
21                     hm.get(stones[i]+step).add(step);
22                     if(step - 1 > 0){
23                         hm.get(stones[i]+step).add(step-1);
24                     }
25                     hm.get(stones[i]+step).add(step+1);
26                 }
27             }
28         }
29         return false;
30     }
31 }