LeetCode #403 - Frog Jump

最新推荐文章于 2021-04-29 23:16:09 发布

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本文详细解析了一道经典的算法题目——青蛙过河。题目要求判断青蛙能否通过一系列石头跳到河对岸，每一步跳跃的距离受到限制。文章通过示例展示了如何使用动态规划的思想解决这一问题，提供了一个C++实现的解决方案。

题目描述：

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.

class Solution {
public:
    bool canCross(vector<int>& stones) {
        int n=stones.size();
        unordered_map<int,set<int>> hash; // 表示从某一块石头的位置上可以跳多少步
        // 必须先初始化，之后一旦发现一个pos不在hash中可以直接跳过
        for(int i=0;i<n;i++) hash[stones[i]]={};
        hash[0].insert(1);
        for(int i=0;i<stones.size();i++)
        {
            for(auto step:hash[stones[i]])
            {
                int nextPos=stones[i]+step;
                if(nextPos==stones.back()) return true;
                // nextPos在hash中存在，说明它是一块石头的位置
                if(hash.count(nextPos)>0) 
                {
                    hash[nextPos].insert(step);
                    hash[nextPos].insert(step+1);
                    if(step>1) hash[nextPos].insert(step-1);
                }
            }
        }
        return false;
    }
};