LeetCode 403 Frog Jump (dp)

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本文探讨了LeetCode上的经典问题“青蛙过河”，通过两种不同的算法实现方式来解决这一问题。一种是基于数组记录每一步可能的位置，另一种则是利用哈希集合优化查找效率。文章详细解释了每种方法的实现思路及代码细节。

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.

题目链接：https://leetcode.com/problems/frog-jump/

题目分析：裸的n^2做法400ms+，lastPos[i]存到到i位置前可能的步长

public class Solution {
    public boolean canCross(int[] stones) {
        int len = stones.length;
        if (len == 2 && stones[1] - stones[0] > 1) {
            return false;
        }
        Set[] lastPos = new Set[len + 1];
        for (int i = 1; i < len; i ++) {
            lastPos[i] = new HashSet<>();
        }
        lastPos[1].add(1);
        for (int i = 2; i < len; i ++) {
            for (int j = 1; j < i; j ++) {
                if (lastPos[j].size() > 0) {
                    int dist = stones[i] - stones[j];
                    if (lastPos[j].contains(dist) || lastPos[j].contains(dist - 1) || lastPos[j].contains(dist + 1)) {
                        lastPos[i].add(dist);
                    }
                }
            }
        }
        return lastPos[len - 1].size() > 0;
    }
}

更好地做法是对于每个位置记录当前位置能向前走的距离，用一个map套set实现，90ms+

public class Solution {
    public boolean canCross(int[] stones) {
        int len = stones.length;
        Map<Integer, HashSet<Integer>> map = new HashMap<>();
        for (int i = 0; i < len; i ++) {
            map.put(stones[i], new HashSet<>());
        }
        map.get(0).add(1);
        for (int i = 0; i < len - 1; i ++) {
            for (int step : map.get(stones[i])) {
                int to = step + stones[i];
                if (to == stones[len - 1]) {
                    return true;
                }
                HashSet<Integer> tmp = map.get(to);
                if (tmp != null) {
                    tmp.add(step);
                    if (step > 1) {
                        tmp.add(step - 1);
                    }
                    tmp.add(step + 1);
                }
            }
        }
        return false;
    }
}