Frog Jump - 非递归实现青蛙过河问题

问题描述

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 has just jumped 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.

问题分析：

方案1：可以使用递归实现，但是会有大量的重复计算，时间复杂度高

方案2：采用宽度优先遍历。因为青蛙仅能向前跳，所以我们可以为每一个石头维护一个set，里面记录可以到达这块石头的所有jump值。我们只需要顺序往前找可以到达的石头，然后将可以从这块石头到达的后续石头加入结果集中。这样只要最后一块石头被加入结果集，就可以认为能到达。代码如下：

class Solution{
public: 
    bool canCross(vector<int> &stones)
    {
    
        vector<set<int> > tmp_rlt_arr(stones.size());
        if (stones[1] != 1) return false;
        tmp_rlt_arr[1].insert(1);


        for (int pos = 1; pos < tmp_rlt_arr.size(); ++pos)
        {
            set<int> &tmp_rlt = tmp_rlt_arr[pos];
            
            for (set<int>::iterator it  = tmp_rlt.begin(); it != tmp_rlt.end(); ++it)
            {
                if (pos == tmp_rlt_arr.size() - 1) return true;
    
                int jump = *it;
                int next_pos;
                //jump -1
                if (jump > 1)
                {
                    next_pos = find_next_pos(stones, pos, stones[pos] + jump -1);
                    if (next_pos > pos) 
                    {
                        tmp_rlt_arr[next_pos].insert(stones[next_pos] - stones[pos]);
                    }
                }
    
                //jump
                next_pos = find_next_pos(stones, pos, stones[pos] + jump);
                if (next_pos > pos) 
                {
                    tmp_rlt_arr[next_pos].insert(stones[next_pos] - stones[pos]);
                }
                //jump +1
                next_pos = find_next_pos(stones, pos, stones[pos] + jump + 1);
                if (next_pos > pos)
                {
                    tmp_rlt_arr[next_pos].insert(stones[next_pos] - stones[pos]);
                }
            }
        }
        return false;
    }
private:
    //binary search next pos
    inline int find_next_pos(vector<int> &stones, int cur_pos, int val)
    {
        int l = cur_pos + 1;
        int r = stones.size() - 1;
        while(l <= r)
        {
            int mid = (l + r)/2;
            if (val > stones[mid])
            {
                l = mid + 1;
            }else if (val < stones[mid])
            {
                r = mid -1;
            }else 
            {
                return mid;
            }
        }
        return -1;
    }
};