Frog Jump

题目来源
青蛙跳的题目，河中有一堆石头，问青蛙能不能跳过河。
我一开始想着深度搜索遍历，代码如下：

class Solution {
public:
    bool canCross(vector& stones) {
        unordered_set sets;
        for (auto stone : stones)
            sets.insert(stone);
        return dfs(0, 1, sets, stones.back());
    }
    
    bool dfs(int cur, int jump, unordered_set& sets, int des)
    {
        if (jump == 0)
            return false;
        cur += jump;
        if (cur == des)
            return true;
        if (sets.count(cur) == 0)
            return false;
        if (dfs(cur, jump-1, sets, des) || dfs(cur, jump, sets, des) 
            || dfs(cur, jump+1, sets, des))
            return true;
        return false;
    }
};

没错…有超时了…
然后想了想，应该是DP，存储一下某个位置某个jump能不能到达，代码如下：

class Solution {
public:
    bool canCross(vector& stones) {
        unordered_set sets;
        for (auto stone : stones)
            sets.insert(stone);
        map, bool> canReach;
        return dfs(0, 1, sets, stones.back(), canReach);
    }
    
    bool dfs(int cur, int jump, unordered_set& sets, int des, map, bool>& canReach)
    {
        if (canReach.count(make_pair(cur, jump)) != 0)
            return canReach[make_pair(cur, jump)];
        if (jump == 0)
            return false;
        cur += jump;
        if (cur == des)
            return true;
        if (sets.count(cur) == 0)
            return false;
        if (dfs(cur, jump-1, sets, des, canReach) || dfs(cur, jump, sets, des, canReach) 
            || dfs(cur, jump+1, sets, des, canReach)) {
                canReach[make_pair(cur, jump)] = true;
                return true;
            }
        canReach[make_pair(cur, jump)] = false;
        return false;
    }
};

然后AC了，感觉爽爽的。不过看了下运行时间，还是比较慢的…
简洁很多的代码，本质上和我的做法是一样的，不过我是把每个都试过，而这个解法是只试石头位置，所以可以剪掉很多
代码如下：

class Solution {
public:
    bool canCross(vector& stones) {
        map, bool> dp;
        return canCross(stones, 0, 0, dp);
    }
    
    bool canCross(vector& stones, int pos, int k, map, bool>& dp)
    {
        if (dp.count(make_pair(pos, k)) != 0)
            return dp[make_pair(pos, k)];
        for (int i=pos+1; i k + 1) {
                dp[make_pair(pos, k)] = false;
                return false;
            }
            if (canCross(stones, i, gap, dp)) return dp[make_pair(pos, k)] = true;
        }
        return pos == stones.size() - 1;
    }
};