403. 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.

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Seen this question in a real interview before?

一直青蛙跳石头，假设前面的石头跳k步调到当前石头，那么当前石头可以跳k，k-1，k+1步，问青蛙能否跳到最后一个石头。

思路1：使用哈希表存储每个石头可以跳的步数，从第一个石头开始，计算后面可到石头的可跳步数。使用了map<int,set<int>>

class Solution {public:    bool canCross(vector<int>& stones) {        if(stones.size()==0) return true;        unordered_map<int,unordered_set<int>> p;        p[0].insert(1);        for(int i=1;i<stones.size();i++)        {            p[stones[i]].insert(0);        }        for(int i=0;i<stones.size();i++)        {            int curS=stones[i];//当前石头的位置,从当前位置看能跳到的位置            for(int step:p[curS])//当前石头存储的可跳步数            {                if(step==0) continue;                int reach=step+curS;//某一个可跳步数跳到的下一位置                if(reach==stones[stones.size()-1]) return true;                if(!p[reach].empty())//如果这个达到的位置上有石头                    {                    p[reach].insert(step);                    if(step>1) p[reach].insert(step-1);                    p[reach].insert(step+1);                }            }        }        return false;    }};