403 Frog Jump

403. Frog Jump

1. Question

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 waskunits, then its next jump must be eitherk- 1,k, ork+ 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 < 2 ^ 31.

• 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.

2. Implementation

(1) HashMap

思路:

利用HashMap, key是stone的位置，value是set of steps， 表示从在该stone位置前面的stone，到达该stone位置所需的steps。比如对于输入数组[0,1,3,5,6,8,12,17]， HashMap应该是: {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], 17=[]}, 6 = [1, 2, 3, 4] 是因为在位置3上的石头可以通过步数3到达6， 根据题意， 在6位置上可以走的步数有 {3 - 1, 3, 3 + 1} 即 {2, 3, 4}, 同理在位置5上，可以通过步数1到达位置6， 在6上可以走的步数有 {1, 2} (步数为0没有意义)， 综上, 6 = [1, 2, 3, 4].

如果当前石头的位置加上可走的步数等于最后一个石头的位置则返回true，否则扫完整个数组(跳过最后一个石头位置)后都无法到达石头最后一个位置，则返回false

class Solution {
    public boolean canCross(int[] stones) {
        if (stones == null || stones.length == 0) {
            return false;
        }

        Map<Integer, Set<Integer>> map = new HashMap();

        for (int stone : stones) {
            map.put(stone, new HashSet());
        }

        map.get(0).add(1);

        int n = stones.length;
        for (int i = 0; i < n - 1; i++) {
            int stone = stones[i];

            for (int step : map.get(stone)) {
                int reach = step + stone;

                if (reach == stones[n - 1]) {
                    return true;
                }

                Set<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;
    }
}

3. Time & Space Complexity

HashMap: 时间复杂度O(n ^ 2), 空间复杂度O(n)