LeetCode题解(Week 8):403. Frog Jump

最新推荐文章于 2020-03-08 00:06:35 发布

Richard_Che

最新推荐文章于 2020-03-08 00:06:35 发布

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本文探讨了一只青蛙如何跨越河流的经典问题。青蛙从第一块石头出发，每一步只能向前跳，且跳跃的距离受到前一步跳跃距离的限制。文章提供了一个算法解决方案，利用动态规划的方法判断青蛙是否能成功到达最后一块石头。

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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.

中文大意

经典的青蛙过河问题，有一个数组a={a1,a2,a3…,an}，其中a1 = 0，一只青蛙从a1出发，第一步走1个单位，青蛙过河有这样的规律：第k-1次如果跳了n个单位，那么第k次就只能跳n,n-1,n+1个单位，并且只能向前跳，问青蛙能否到达对岸（也就是跳到an）

题解

class Solution {
public:
    bool canCross(vector<int>& stones) {
        int n = stones.size();
        unordered_map<int, unordered_set<int>> dp;
        dp[0].insert(0);

        int maxdist = 0;
        for(int i = 1 ;i < n;i++)
        {
            for(int j = 0;j<i;j++)
            {
                int currdist = stones[i] - stones[j];
                //如果跳maxdist+1跳不到现在的这个石头，那就不用考虑j这块石头了
                if(currdist > maxdist+1) continue;
                if(dp[j].count(currdist)||dp[j].count(currdist-1)||dp[j].count(currdist+1))
                {
                    dp[i].insert(currdist);
                    maxdist = max(currdist,maxdist);
                }
            }
        }

        return dp[n-1].size() > 0;
    }
};