推 UsadaBeguora: 露醬最強 08/27 17:40
→ sustainer123: 大師 08/27 18:00
https://leetcode.com/problems/frog-jump/description/
403. Frog Jump
給你一個陣列 stones[]，stones[i] 表示石頭的座標，青蛙會從 stones[0] 開始跳，
第一次跳 1 步，後面可以跳 k, k + 1, k - 1 步（k 為上一次跳的步數），青蛙只能
走在石頭上，求出青蛙是否可以跳到最後一個石頭的位置。
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.
思路：
1.從第一個石頭往後跳 1 步並標記目的地石頭為可到達，之後遍歷所有的石頭，如果這
個石頭是可到達的就往後跳 k, k+1, k-1 步，如果目的地是石頭也標記為可到達。
2.最後檢查最後一顆石頭是不是可到達即可。
Java Code：
----------------------------------------------------
class Solution {
public boolean canCross(int[] stones) {
Map<Integer, Set<Integer>> map = new HashMap<>();
for (int stone : stones) {
map.put(stone, new HashSet<>());
}
map.computeIfPresent(stones[0] + 1, (k, v) -> {
v.add(1);
return v;
});
for (int i = 1; i < stones.length; i++) {
Set<Integer> set = new HashSet<>(map.get(stones[i]));
for (int step : set) {
if (map.containsKey(stones[i] + step)) {
map.get(stones[i] + step).add(step);
}
if (map.containsKey(stones[i] + step + 1)) {
map.get(stones[i] + step + 1).add(step + 1);
}
if (map.containsKey(stones[i] + step - 1)) {
map.get(stones[i] + step - 1).add(step - 1);
}
}
}
return map.get(stones[stones.length - 1]).size() > 0;
}
}
----------------------------------------------------
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https://i.imgur.com/4nfnn6f.jpg
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