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.

解法

这一道题我一开始使用回溯法超时了，所以考虑使用了DP。

依旧是一个二维数组dp[][]，数组建立的大小为 stones序列的长度 X 长度

dp数组的含义  dp[a][b]

a:   表示现在在第几块石头

b：表示走到这块石头之前走了几格（因为下一步能走的长度取决于之前走的长度）

dp的值为true or false 表示当前这种情况是否存在

 

由于本题的题意设置，不用担心后来走的长度会超过dp的宽度，因为第一步最多走1，第二步最多走2，。。第n步 最多走n

也就是数组中的b不会超过stones序列的长度

算法流程：

由第一个石头开始，试探0或1步，如果能抵达第二块，在第二块根据之前的步长继续试探，直到最后

我们只需要检查最后一行的数组是否存在true就可以知道是否有解。

 

代码：

class Solution {
    
     public boolean canCross(int[] stones) {
        boolean[][] dp = new boolean[stones.length][stones.length];
        dp[0][0] = true;
        for(int i = 0; i < stones.length; i++){
            for(int prevGap = 0; prevGap <= i; prevGap++){
                if(dp[i][prevGap]){   //当前有可行方案
                    for(int next = i + 1; next < stones.length && stones[next] - stones[i] <= prevGap + 1 ; next++){
                        if(prevGap - 1 > stones[next] - stones[i]) continue; //石头间距太小，无法走到，跳过
                        int gap = stones[next] - stones[i];   //计算走的步长
                        dp[next][gap] = true;    //记录
                    }
                }
            }
        }
        for(boolean b : dp[dp.length - 1])     //找最后一行是否存在true
            if(b) return true;
        return false;
    }
}