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;
} }

LeetCode403. Frog Jump的更多相关文章

  1. [Swift]LeetCode403. 青蛙过河 | Frog Jump

    A frog is crossing a river. The river is divided into x units and at each unit there may or may not ...

  2. [LeetCode] Frog Jump 青蛙过河

    A frog is crossing a river. The river is divided into x units and at each unit there may or may not ...

  3. Frog Jump

    A frog is crossing a river. The river is divided into x units and at each unit there may or may not ...

  4. Leetcode: Frog Jump

    A frog is crossing a river. The river is divided into x units and at each unit there may or may not ...

  5. [leetcode]403. Frog Jump青蛙过河

    A frog is crossing a river. The river is divided into x units and at each unit there may or may not ...

  6. [LeetCode] 403. Frog Jump 青蛙跳

    A frog is crossing a river. The river is divided into x units and at each unit there may or may not ...

  7. div 3 frog jump

    There is a frog staying to the left of the string s=s1s2…sn consisting of n characters (to be more p ...

  8. [leetcode] 403. Frog Jump

    https://leetcode.com/contest/5/problems/frog-jump/ 这个题目,还是有套路的,之前做过一道题,好像是贪心性质,就是每次可以跳多远,最后问能不能跳到最右边 ...

  9. 第十七周 Leetcode 403. Frog Jump(HARD) 线性dp

    leetcode403 我们维护青蛙从某个石头上可以跳那些长度的距离即可 用平衡树维护. 总的复杂度O(n^2logn) class Solution { public: bool canCross( ...

随机推荐

  1. NIN (Network In Network)

    Network In Network 论文Network In Network(Min Lin, ICLR2014). 传统CNN使用的线性滤波器是一种广义线性模型(Generalized linea ...

  2. 一、linux学习之centOS系统安装(VMware下安装)

    一.下载 这个真的没有什么技术含量,也不附下载连接了.这里需要说明的是,其实在VMware下安装centOS是非常简单的,但是这里我要纪录的是在PC上安装centOS,之所以跟标题有出入是因为为了纪录 ...

  3. 拥抱Service Fabric —— 目录

    理解分布式 经典分布式系统设计 云时代分布式系统演进 Service Fabric基础概念 Node, Application, Service, Partition/Replicas Partiti ...

  4. Python高手之路【四】python函数装饰器,迭代器

    def outer(func): def inner(): print('hello') print('hello') print('hello') r = func() print('end') p ...

  5. ORM choice字段 如何在页面上显示值

    核心:obj.get_字段名_display 1.定义module 数据结构: class msg(models.Model): choice = ( (1, '技术部'), (2, '行政'), ( ...

  6. python---await/async关键字

    推文:玩转 Python 3.5 的 await/async 首先看正常的两个函数之间的执行 def func1(): print("func1 start") print(&qu ...

  7. centos7.2 rabbitmq3.6.2源码部署

    1.安装所有依赖包yum install -y gcc ncurses ncurses-base ncurses-devel ncurses-libs ncurses-static ncurses-t ...

  8. 移动端UI

    mui:http://dev.dcloud.net.cn/mui/ saltUI:https://salt-ui.github.io

  9. POJ 1741 Tree 求树上路径小于k的点对个数)

                                                                                                 POJ 174 ...

  10. jetty 热部署

    1,在pom.xml文件中配置jetty插件的参数:scanIntervalSeconds <plugin> <groupId>org.mortbay.jetty</gr ...