Frog Jump

题目地址：
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.

解题思路：
这道题就是用dfs可以解决，路径的话可以用list来存，然后记得加多少然后dfs完了减多少。

代码：

public class FrogJump {

//    public boolean canCross(int[] stones) {////    }
    public List<List<Integer>> canCross(int[] stones) {
        List<List<Integer>> rst = new ArrayList<>();
        if(stones == null || stones.length == 0){
            return rst;
        }
        List<Integer> list = new ArrayList<>();
        if(stones.length == 1){
            list.add(0);
            rst.add(list);
            return rst;
        }
        dfs(0, 0, list, rst, stones);
        return rst;
    }

    private void dfs(int start, int lastJump, List<Integer> list, List<List<Integer>> rst, int[] stones) {
        if(start == stones.length - 1){
            list.add(start);
            rst.add(new ArrayList<>(list));
            list.remove(list.size() - 1);
            return;
        }
        list.add(start);
        if(start == 0){
            if(stones[1] != 1){
                return;
            }
            else{
                dfs(1, 1, list, rst, stones);
            }
        }
        else{
            for(int i = lastJump - 1; i <= lastJump + 1; i++){
                int temp = canJump(start, i, stones);
                if(temp != -1){
                    dfs(temp, i, list, rst, stones);
                }
            }
        }
        list.remove(list.size() - 1);
    }

    private int canJump(int start, int jump, int[] stones) {
        int i = start + 1;
        while(i <= stones.length - 1){
            if(stones[start] + jump == stones[i]){
                return i;
            }
            else if(stones[start] + jump < stones[i]){
                break;
            }
            i++;
        }
        return -1;
    }

    public static void main(String[] args){
        FrogJump frogJump = new FrogJump();
        int[] stones = {0, 1, 3, 5, 6, 8, 9, 12, 16};
//        int[] stones = {0, 1, 3, 5, 6};        List<List<Integer>> rst = frogJump.canCross(stones);
        System.out.println(rst);
    }

}

这里打印路径其实可以用类似于动态规划的解法。存一下到每个点的路径。

其实这道题的原题是return true or false：

public boolean canCross(int[] stones) {
        if (stones == null || stones.length == 0) {return false;}
        int n = stones.length;
        if (n == 1) {return true;}
        if (stones[1] != 1) {return false;}
        if (n == 2) {return true;}
        int last = stones[n - 1];
        HashSet<Integer> hs = new HashSet();
        for (int i = 0; i < n; i++) {
            if (i > 3 && stones[i] > stones[i - 1] * 2) {return false;} // The two stones are too far away. 
            hs.add(stones[i]);
        }
        return canReach(hs, last, 1, 1);
    }
    
    private boolean canReach(HashSet<Integer> hs, int last, int pos, int jump) {
        if (pos + jump - 1 == last || pos + jump == last || pos + jump + 1 == last) {
            return true;
        }
        if (hs.contains(pos + jump + 1)) {
            if (canReach(hs, last, pos + jump + 1, jump + 1)) {return true;}
        }
        if (hs.contains(pos + jump)) {
            if (canReach(hs, last, pos + jump, jump)) {return true;}
        }
        if (jump > 1 && hs.contains(pos + jump - 1)) {
            if (canReach(hs, last, pos + jump - 1, jump - 1)) {return true;}
        }
        return false;
    }