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.
原题链接
解题思想
小青蛙来过河，河被分为x个单元，每个单元都有两种情况，有石头和没石头。青蛙可以跳到石头上但不能跳进水里。下面给出的字符串由低到高表示石头所在的单元。判断青蛙能跳到最后一块石头上吗？
初始时，青蛙在第一块石头上，而且第一次只跳一个单元。
如果青蛙最后一次跳了k步，那么它下一步只能跳k-1或k+1步。注意，青蛙只能往前跳。

分析：使用map纪录青蛙在当前的石头上所能跳的步数，如果最后能到达最后一块石头，返回true.
对于：Example 1:
[0,1,3,5,6,8,12,17]
由题意知，0={1},1={1,2},3={1,2,3},5={1,2,3},6={1,2,3,4},8={1,2,3,4},12={3,4,5}。
每跳到一块石头上，就对后面能到达的石头的步数集合进行更新。
这里map的key是石头下标，第二部分用set表示步数集合。

public:
    bool canCross(vector<int>& stones) {
        int n=stones.size();
       map<int,set<int>> mp;
       set<int> t;
       for(int i=0;iint,set<int>>(stones[i],t));
       //每个石头都有set

       mp[0].insert(1);//初始化，0号石头的步数集合{1}
       for(int i=0;i1;i++)
       {
           for(auto step:mp[stones[i]])
           {
              int reach=stones[i]+step;
              if(reach==stones[n-1])
                return true;
             //每到达一块石头上，更新下对应的set
             if(mp.find(reach)!=mp.end())
             {
             mp[reach].insert(step);
             mp[reach].insert(step+1);
             if(step-1>0) mp[reach].insert(step-1);
             }
           }    
       }
       return false;
    }