403. Frog Jump

最新推荐文章于 2021-04-29 23:16:09 发布

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最新推荐文章于 2021-04-29 23:16:09 发布

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文章标签： c/c++

原文链接：http://www.cnblogs.com/ruruozhenhao/p/10403693.html

本文探讨了一个经典的动态规划问题——青蛙过河。通过分析青蛙如何从一系列石头上跳跃过河，我们详细解释了如何使用动态规划算法来确定青蛙是否能够成功到达对岸。文章提供了一个C++实现的示例，展示了如何通过维护一个映射来跟踪青蛙在每个石头上的可能跳跃长度，从而找到解决方案。

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.

Approach #1: DP. [C++]

class Solution {
public:
    bool canCross(vector<int>& stones) {
        int size = stones.size();
        map<int, set<int>> mp;
        for (int i = 0; i < size; ++i) mp[stones[i]] = {};
        
        mp[0] = {0};
        for (int i = 0; i < size; ++i) {
            for (int lastStep : mp[stones[i]]) {
                for (int nextStep = lastStep-1; nextStep <= lastStep+1; ++nextStep) {
                    int idx = nextStep + stones[i];
                    if (nextStep > 0 && mp.count(idx)) {
                        mp[idx].insert(nextStep);
                    }
                }
            }
        }
        
        return !mp[stones[size-1]].empty();
    }
};

Analysis:

mp[stones[i]] : represent how many lastStep that we can get to stones[i].

mp[0] = {0} : initialite the flog's position at the first stone.

转载于:https://www.cnblogs.com/ruruozhenhao/p/10403693.html