README.md

207. Course Schedule

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

• For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]] Output: true Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. 

Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]] Output: false Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible. 

Constraints:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= 5000
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• All the pairs prerequisites[i] are unique.

Solutions (Rust)

1. Solution

use std::collections::HashMap;implSolution{pubfncan_finish(num_courses:i32,prerequisites:Vec<Vec<i32>>) -> bool{letmut indegrees = vec![0; num_courses asusize];letmut follows = HashMap::new();letmut courses = vec![];for prerequisite in prerequisites { indegrees[prerequisite[0]asusize] += 1; follows .entry(prerequisite[1]asusize).or_insert(vec![]).push(prerequisite[0]asusize);}for b in0..num_courses asusize{if indegrees[b] == 0{ courses.push(b);}}whileletSome(b) = courses.pop(){for a in follows.remove(&b).unwrap_or(vec![]){ indegrees[a] -= 1;if indegrees[a] == 0{ courses.push(a);}}} indegrees.iter().all(|&x| x == 0)}}