403. Frog Jump(Leetcode每日一题-2021.04.29)--抄答案
2021/4/29 10:26:08
本文主要是介绍403. Frog Jump(Leetcode每日一题-2021.04.29)--抄答案,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
Problem
A frog is crossing a river. The river is divided into some number of 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 can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.
If the frog’s last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.
Constraints:
- 2 <= stones.length <= 2000
- 0 <= stones[i] <= 2^31 - 1
- stones[0] == 0
Example1
Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: 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.
Example2
Input: stones = [0,1,2,3,4,8,9,11]
Output: false
Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.
Solution
记忆化搜索
class Solution { public: vector<vector<int>> f; unordered_map<int, int> hash; vector<int> stones; int dp(int i, int j) { if (f[i][j] != -1) return f[i][j]; f[i][j] = 0; for (int k = max(1, j - 1); k <= j + 1; k ++ ) { if (hash.count(stones[i] - k)) { int p = hash[stones[i] - k]; if (dp(p, k)) { f[i][j] = 1; break; } } } return f[i][j]; } bool canCross(vector<int>& _stones) { stones = _stones; int n = stones.size(); f = vector<vector<int>>(2*n,vector<int>(2*n,-1)); for (int i = 0; i < n; i ++ ) hash[stones[i]] = i; f[0][1] = 1; for (int i = 0; i < n; i ++ ) if (dp(n - 1, i)) return true; return false; } };
这篇关于403. Frog Jump(Leetcode每日一题-2021.04.29)--抄答案的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!
- 2024-05-15PingCAP 黄东旭参与 CCF 秀湖会议,共探开源教育未来
- 2024-05-13PingCAP 戴涛:构建面向未来的金融核心系统
- 2024-05-09flutter3.x_macos桌面os实战
- 2024-05-09Rust中的并发性:Sync 和 Send Traits
- 2024-05-08使用Ollama和OpenWebUI在CPU上玩转Meta Llama3-8B
- 2024-05-08完工标准(DoD)与验收条件(AC)究竟有什么不同?
- 2024-05-084万 star 的 NocoDB 在 sealos 上一键起,轻松把数据库编程智能表格
- 2024-05-08Mac 版Stable Diffusion WebUI的安装
- 2024-05-08解锁CodeGeeX智能问答中3项独有的隐藏技能
- 2024-05-08RAG算法优化+新增代码仓库支持,CodeGeeX的@repo功能效果提升