标签：理解 order bubuko forward 遍历 sort int 内存 fir
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:
[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.
题意：
一只不会游泳的青蛙，从位置0th unit(即我画的图中的第0号石头)出发开始跳，
若前一次它跳了K units，
这次就只能跳 K-1 或者 K 或者 K+1 units，
问青蛙是否能跳到对岸。
思路：
以[0,1,3,5,6,8,12,17]为例自己画了个图，我把units理解为直尺上的刻度，更直观理解题意。
用HashMap套一个set，set内存放last jump的units
首先，遍历一遍给定数组stones = {0,1,3,5,6,8,12,17}， 存入所有给定刻度，初始化HashMap。
Integer(当前刻度) 0 1 3 5 6 8 12 17
Set(last jump units) [] [] [] [] [] [] [] []
然后，初始化0刻度对应的Set为[0]。
Integer(当前刻度) 0 1 3 5 6 8 12 17
Set(last jump units) [0] [] [] [] [] [] [] []
最后，遍历一遍给定数组stones = {0,1,3,5,6,8,12,17}
根据题意，青蛙的下一跳(nextJump)范围取决于上一跳(lastJump)
遍历Set里上一跳的取值
Integer(当前刻度) 0 1 3 5 6 8 12 17 Set(last jump units) [0] [] [] [] [] [] [] [] lastJump 0 nextJump -1 0 1
查找map.containsKey(下一跳的刻度)，即 (当前刻度 + nextJump Range)
Integer(当前刻度) 0 1 3 5 6 8 12 17 Set(last jump units) [0] [1] [2] [2] [3,1] [3,2] [4] [5] lastJump 0 1 2 2 3 1 3 2 4 5 nestJump Range: (1)lastJump-1 -1 0 1 1 2 0 2 1 3 (2)lastJump 0 1 2 2 3 1 3 2 4 (3)lastJump+1 1 2 3 3 4 2 4 3 5 map.containsKey 0-1? 1+0? 3+1? 5+1? 6+2? 6+0? 8+2? 8+1? 12+3? 0+0? 1+1? 3+2? 5+2? 6+3? 6+1? 8+3? 8+2? 12+4? 0+1? 1+2? 3+3? 5+3? 6+4? 6+2? 8+4? 8+3? 12+5?
代码：
1 public boolean canCross(int[] stones) { 2 if (stones.length == 0) return false; 3 HashMap<Integer, HashSet<Integer>> map = new HashMap<>(); 4 for (int i = 0; i < stones.length; i++) { 5 map.put(stones[i], new HashSet<>()); 6 } 7 8 map.get(0).add(0); 9 for (int i = 0; i < stones.length; i++) { 10 for (int lastJump : map.get(stones[i])) { 11 for (int nextJump = lastJump - 1; nextJump <= lastJump + 1; nextJump++) { 12 if (nextJump > 0 && map.containsKey(stones[i] + nextJump)) { 13 map.get(stones[i] + nextJump).add(nextJump); 14 } 15 } 16 } 17 } 18 return !map.get(stones[stones.length - 1]).isEmpty(); 19 }
标签：理解 order bubuko forward 遍历 sort int 内存 fir
原文地址：https://www.cnblogs.com/liuliu5151/p/9114677.html
