原题链接在这里：https://leetcode.com/problems/frog-jump/description/
题目：
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.
题解：
看青蛙是否能正好跳到最后一块石头上.
DP问题，需要保留的历史信息是在当前这块石头上能向前跳多少步.
状态转移时，若在当前石头向前跳n步时刚好能落到后面的石头上，那么后面石头就能能跳n-1, n , n+1步，把这三种情况加到后面那块石头的存储信息里.
答案是看中间有没有正好跳到最后一块石头上的可能.
Time Complexity: O(n^2). n = stones.length. 最坏情况每个step都能跳到，那么对于后面的石头就有接近3n种可向前跳的步数.
Space: O(n).
AC Java:
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 }
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/8319740.html