[LeetCode] 403. Frog Jump 青蛙跳

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.

一只青蛙要跳过一条河，河被分成了x个单元，每个单元里有或者没有石头，青蛙可以跳到石头上，不能跳到河里。 给一个石头位置的数组n，按升序排列， 判断是否青蛙可以跳到最后一个石头上，青蛙从第一个石头开始跳。如果最后一跳是k单元， 下一个跳发是k-1, k, 或者k+1， 青蛙只能往前跳。

解法：DP + Hash table

Python:

# DP with hash table
class Solution(object):
    def canCross(self, stones):
        """
        :type stones: List[int]
        :rtype: bool
        """
        if stones[1] != 1:
            return False

        last_jump_units = {s: set() for s in stones}
        last_jump_units[1].add(1)
        for s in stones[:-1]:
            for j in last_jump_units[s]:
                for k in (j-1, j, j+1):
                    if k > 0 and s+k in last_jump_units:
                        last_jump_units[s+k].add(k)
        return bool(last_jump_units[stones[-1]])　　

C++:

class Solution {
public:
    bool canCross(vector<int>& stones) {
        unordered_map<int, bool> m;
        return helper(stones, 0, 0, m);
    }
    bool helper(vector<int>& stones, int pos, int jump, unordered_map<int, bool>& m) {
        int n = stones.size(), key = pos | jump << 11;
        if (pos >= n - 1) return true;
        if (m.count(key)) return m[key];
        for (int i = pos + 1; i < n; ++i) {
            int dist = stones[i] - stones[pos];
            if (dist < jump - 1) continue;
            if (dist > jump + 1) return m[key] = false;
            if (helper(stones, i, dist, m)) return m[key] = true;
        }
        return m[key] = false;
    }
};

C++:

class Solution {
public:
    bool canCross(vector<int>& stones) {
        unordered_map<int, unordered_set<int>> m;
        vector<int> dp(stones.size(), 0);
        m[0].insert(0);
        int k = 0;
        for (int i = 1; i < stones.size(); ++i) {
            while (dp[k] + 1 < stones[i] - stones[k]) ++k;
            for (int j = k; j < i; ++j) {
                int t = stones[i] - stones[j];
                if (m[j].count(t - 1) || m[j].count(t) || m[j].count(t + 1)) {
                    m[i].insert(t);
                    dp[i] = max(dp[i], t);
                }
            }
        }
        return dp.back() > 0;
    }
};

原文地址：https://www.cnblogs.com/lightwindy/p/9691934.html