【LeetCode】403. Frog Jump

题目：
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.
分析：
这道题是用动态规划求解。每一个石头都用集合set来存从该石头可能跳跃的距离，再用一个一维数组dis来存储每块石头可以跳跃的最长距离。初始化第0个石头可以跳的最长距离为1.
遍历整个数组，通过dis来找跳跃距离可以超过或等于当前石头的石头j，然后从j开始找有没有符合要求的石头。若有，则将该距离添加到该石头对应的set中并更新dis。
最后判断最后一个石头的dp的长度，若为0，表示不能到达最后一个石头，返回false，不然返回true。

代码：

class Solution {
public:
    bool canCross(vector<int>& stones) {
        vector<set<int>> dp;
        vector<int> dis;
        for(int i = 0;i<stones.size();i++)
        {
            dp.push_back(set<int>());
            dis.push_back(0);
        }
        dp[0].insert(1);
        int j;
        for(int i = 1;i<stones.size();i++)
        {
            for(j = 0;j<i;j++)
            {
                if(dis[j]+1>=stones[i]-stones[j])
                break;
            }
            for(int p = j;p<i;p++)
            {
                int temp = stones[i]-stones[p];
                if(dp[p].find(temp)!=dp[p].end()||dp[p].find(temp-1)!=dp[p].end()||dp[p].find(temp+1)!=dp[p].end())
                {
                    dp[i].insert(temp);
                    if(temp>dis[i])
                    dis[i] = temp;
                }
            }

        }
        if(dp[stones.size()-1].size()!=0)
        return true;
        else return false;
    }
};