LeetCode每日一题(1857. Largest Color Value in a Directed Graph)

给定一个由n个彩色节点和m条边的有向图，每个节点从0到n-1编号。任务是找到任何有效路径的最大颜色值，颜色值是该路径上出现最频繁的颜色的节点数量。如果图中存在循环，则返回-1。提供的代码实现了一个解决方案，通过计算节点的入度、颜色计数和使用队列进行广度优先搜索来找出最大颜色值路径。

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There is a directed graph of n colored nodes and m edges. The nodes are numbered from 0 to n - 1.

You are given a string colors where colors[i] is a lowercase English letter representing the color of the ith node in this graph (0-indexed). You are also given a 2D array edges where edges[j] = [aj, bj] indicates that there is a directed edge from node aj to node bj.

A valid path in the graph is a sequence of nodes x1 -> x2 -> x3 -> … -> xk such that there is a directed edge from xi to xi+1 for every 1 <= i < k. The color value of the path is the number of nodes that are colored the most frequently occurring color along that path.

Return the largest color value of any valid path in the given graph, or -1 if the graph contains a cycle.

Example 1:

Input: colors = “abaca”, edges = [[0,1],[0,2],[2,3],[3,4]]
Output: 3

Explanation: The path 0 -> 2 -> 3 -> 4 contains 3 nodes that are colored “a” (red in the above image).

Example 2:

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Input: colors = “a”, edges = [[0,0]]
Output: -1

Explanation: There is a cycle from 0 to 0.

Constraints:

n == colors.length
m == edges.length
1 <= n <= 10^5
0 <= m <= 10^5
colors consists of lowercase English letters.
0 <= aj, bj < n

提示基本把解这题的关键点都说清楚了，按提示中的方式来写即可


impl Solution {
    pub fn largest_path_value(colors: String, edges: Vec<Vec<i32>>) -> i32 {
        let colors: Vec<char> = colors.chars().collect();
        let mut counts = vec![vec![0; 26]; colors.len()];
        let mut indegrees = vec![0; colors.len()];
        let edges = edges.into_iter().fold(vec![Vec::new(); colors.len()], |mut l, e| {
            indegrees[e[1] as usize] += 1;
            counts[e[0] as usize][colors[e[0] as usize] as usize - 97] = 1;
            counts[e[1] as usize][colors[e[1] as usize] as usize - 97] = 1;
            l[e[0] as usize].push(e[1] as usize);
            l
        });
        let mut queue: Vec<usize> = indegrees.iter().enumerate().filter_map(|(i, &v)| if v == 0 { Some(i as usize) } else { None }).collect();
        let mut ans = 1;
        let mut visited = 0;
        while let Some(n) = queue.pop() {
            visited += 1;
            ans = ans.max(counts[n].iter().map(|v| *v).max().unwrap());
            for &next in &edges[n] {
                for c in 'a'..='z' {
                    counts[next][c as usize - 97] = counts[next][c as usize - 97].max(counts[n][c as usize - 97] + if colors[next] == c { 1 } else { 0 });
                }
                indegrees[next] -= 1;
                if indegrees[next] == 0 {
                    queue.push(next);
                }
            }
        }
        if visited < colors.len() {
            -1
        } else {
            ans
        }
    }
}