#### 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.
```

题目大意: 给定一组数, 代表河里有石头的位置,如果青蛙从前一个石头跨了k步到现在的石头,那么它下一次只能走k-1,k,或k+1步。

附: 原题目链接

#### 思路

用一个dictionary 来存储到达当前石头的话，所有上一次的可能的步数, 然后根据(k-1,k,k+1)这个规则来决定下一次可以到的石头。

```
class Solution {
func canCross(_ stones: [Int]) -> Bool {
guard !stones.isEmpty else {
return false
}
let stonesSet = Set(stones)
var dict = [Int: Set<Int>]()
dict[0] = [0]
for stone in stones {
if let steps = dict[stone] {
for step in steps {
for possible in (step - 1)...(step + 1) {
if possible > 0 {
let nextLocation = stone + possible
if stonesSet.contains(nextLocation) {
if var locationSteps = dict[nextLocation] {
locationSteps.insert(possible)
dict[nextLocation] = locationSteps
} else {
dict[nextLocation] = [possible]
}
}
}
}
}
}
}
let last = stones.last!
return dict[last] != nil
}
}
```