Leetcode 403. Frog Jump

Description

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.

Solution

题目大意是说青蛙通过跳石头过河,设上一次跳跃的距离为k,则这一次能跳跃的距离为k - 1k或者k + 1,初始在0点,能跳跃的距离为1,让判断青蛙能否过河。

先分析题目应该用什么解法。

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.

从这条描述看得出来,每一次的决策依赖前一次的结果,看起来挺像动态规划的转移方程的。那么我们用动态规划试下。

dp[i][j]代表从i位置跳跃j步,值为true/false,代表能否跳到。

因此初始状态dp[0][1] = true

那么dp数组的第一维大小应当是n,第二维是多大呢?

假设石头序列是0 1 2 3 4 5,则有:

stone : 0 1 2 3 4 5
steps : 1 2 3 4 5 6

在石头0处,我们可以跳跃的最大step为1,在石头1处最大step为2,因此在石头k处,最大step为n+1。

因此有最终代码:

class Solution {
    public boolean canCross(int[] stones) {
        int n = stones.length;
        boolean[][] dp = new boolean[n][n + 1];
        dp[0][1] = true;
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int distance = stones[i] - stones[j];
                if (distance < 0 || distance > n || !dp[j][distance]) {
                    continue;
                }
                dp[i][distance] = true;
                if (distance - 1 >= 0) dp[i][distance - 1] = true;
                if (distance + 1 <= n) dp[i][distance + 1] = true;
                if (i == n - 1) return true;
            }
        }
        return false;
    }
}

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