LeetCode403. 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.

分析

因为每一步可以走的步长都是有限的，所以想到用dfs来遍历所有的可能，不过超时了，还是看了下网友的解答。

利用一个map来代表从stone到step的映射，表示从这个stone出发的下一步能走多少步长，比如说：

[0,1,3,5,6,8,12,17]

{17=[], 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]}

因为到第三个石头只能是第二个石头走两步，所以当我们 到达第三个石头，再向后出发时，下一步的步长只能是1，2，3。当然，对于最后一个石头我们无需计算。

上面的过程就是不断需找某个石头，能由它前面的哪些石头到达，如果按照一定的步数k到达，并且这个步数是上个石头可以走的，那么我们就可以在当前stone所对应的step中填加上k-1，k，k+1。如果我们最后找到了最后一个stone则返回true，并且不需要计算最后的stone所对应的steps。

代码

import java.util.*;

public class LeetCode {

  public boolean canCross(int[] stones) {
    if (stones.length == 0) {
      return true;
    }

    HashMap<Integer, HashSet<Integer>> map = new HashMap<Integer, HashSet<Integer>>(stones.length);
    map.put(0, new HashSet<Integer>());
    map.get(0).add(1);
    for (int i = 1; i < stones.length; i++) {
      map.put(stones[i], new HashSet<Integer>() );
    }

    for (int i = 0; i < stones.length - 1; i++) {
      int stone = stones[i];
      for (int step : map.get(stone)) {
        int reach = step + stone;
        if (reach == stones[stones.length - 1]) {
          return true;
        }
        HashSet<Integer> set = map.get(reach);
        if (set != null) {
          set.add(step);
          if (step - 1 > 0) set.add(step - 1);
          set.add(step + 1);
        }
      }
    }

    return false;
  }

}

原文地址：https://www.cnblogs.com/f91og/p/9721047.html