Frog Jump

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.

虐到心痛的DP题。

一开始DFS果断超时，然后想DP解法，不做cache纪录的话，则每次需要寻找上上一步从哪里来，O(n^3)的复杂度。

1.DP O(n^2)的解法，使用hashmap[i]保存到达i的step集合。之后判断能否从上一步到这一步，直接核查上一步的step集合里是否存在当前step的step－1，step，step＋1.但是该解法Java可过，python超时。

class Solution(object):
    def canCross(self, stones):
        """
        :type stones: List[int]
        :rtype: bool
        """
        if not stones:
            return True
        n = len(stones)
        hashmap = collections.defaultdict(set) 
        if stones[1] > 1:
            return False
        else:
            hashmap[1].add(1)
        for i in xrange(2,n):
            for j in xrange(1,i):
                if len(hashmap[j]) > 0:
                    step = stones[i] - stones[j]
                    if (step in hashmap[j]) or ((step-1) in hashmap[j]) or ((step+1) in hashmap[j]):
                        hashmap[i].add(step)
        return len(hashmap[n-1]) > 0
        

2.DP，但是hashmap[i]保存的是从i出发可以采用的步长。可以采用的步长数目远远小于历史步数的范围，所以这是一个最差O(n^2)，最好接近于O(n)的解法。代码如下：

class Solution(object):
    def canCross(self, stones):
        """
        :type stones: List[int]
        :rtype: bool
        """
        if not stones:
            return True
        n = len(stones)
        hashmap = collections.defaultdict(set) 
        if stones[1] > 1:
            return False
        else:
            hashmap[1].add(1)
        for i in xrange(2,n):
            for j in xrange(1,i):
                if len(hashmap[j]) > 0:
                    step = stones[i] - stones[j]
                    if (step in hashmap[j]) or ((step-1) in hashmap[j]) or ((step+1) in hashmap[j]):
                        hashmap[i].add(step)
        return len(hashmap[n-1]) > 0
        

 

转载于:https://www.cnblogs.com/sherylwang/p/5934155.html