403 Frog Jump

Leetcode

A frog is crossing a river. The river is divided into some number of 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 can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.

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

Example 1:

Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: 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:

Input: stones = [0,1,2,3,4,8,9,11]
Output: false
Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.

最开始尝试用递归做，但递归会超时，因为有太多重复的点尝试了重复的步骤。。。可以用动态规划来做，每个点有一个 set 来记录从上一个陆地过来时走的距离，尝试每个点可以走的距离，如果能走到陆地，则在该陆地的 set 中记录可以通过多少距离走到这个陆地。

public boolean canCross(int[] stones) {
    if(stones[1]-stones[0]!=1) return false;
    int len=stones.length;
    int lastStone=stones[len-1];
    HashMap<Integer, HashSet<Integer>> map=new HashMap<>(len);
    for(int i=1;i<len;++i){
        map.put(stones[i], new HashSet<Integer>());
    }
    map.get(1).add(1);
    for(int i=1;i<stones.length;++i){
        int stone = stones[i];
        for(int step:map.get(stone)){
            if(stone+step==lastStone+1 || stone+step==lastStone || stone+step==lastStone-1) return true;
            if(map.containsKey(stone+step)){
                map.get(stone+step).add(step);
            }
            if(map.containsKey(stone+step+1)){
                map.get(stone+step+1).add(step+1);
            }
            if(step-1>0 && map.containsKey(stone+step-1)){
                map.get(stone+step-1).add(step-1);
            }
        }
    }
    return false;
}