LeetCode每日一题(2246. Longest Path With Different Adjacent Characters)

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them.

Example 1:

Input: parent = [-1,0,0,1,1,2], s = “abacbe”
Output: 3

Explanation: The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned.
It can be proven that there is no longer path that satisfies the conditions.

Example 2:

Input: parent = [-1,0,0,0], s = “aabc”
Output: 3

Explanation: The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

Constraints:

• n == parent.length == s.length
• 1 <= n <= 105
• 0 <= parent[i] <= n - 1 for all i >= 1
• parent[0] == -1
• parent represents a valid tree.
• s consists of only lowercase English letters.

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题不难， 但是思路很难表达出来， 整体来说就是对于任何一个节点 node， 我们通过经过该节点的两个最长的连续路径的来更新最终结果, 同时返回最长的连续路径来供父节点来计算新的值。

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use std::collections::BinaryHeap;

impl Solution {
    fn rc(children: &Vec<Vec<usize>>, chars: &Vec<char>, i: usize, ans: &mut i32) -> i32 {
        if i == 100000 {
            return 0;
        }
        let sc = chars[i];
        let mut equals = BinaryHeap::new();
        let mut not_equals = BinaryHeap::new();
        for &c in &children[i] {
            let length = Solution::rc(children, chars, c, ans);
            if c != 100000 {
                if chars[c] != sc {
                    not_equals.push(length);
                    continue;
                }
                equals.push(length);
            }
        }
        let neq_first = if let Some(len) = not_equals.pop() {
            len
        } else {
            0
        };
        let neq_second = if let Some(len) = not_equals.pop() {
            len
        } else {
            0
        };
        *ans = (*ans).max(neq_first + neq_second + 1);
        if let Some(len) = equals.pop() {
            *ans = (*ans).max(len);
        }
        neq_first + 1
    }
    pub fn longest_path(parent: Vec<i32>, s: String) -> i32 {
        let chars: Vec<char> = s.chars().collect();
        let mut children = vec![Vec::new(); parent.len()];
        for (i, p) in parent.into_iter().enumerate() {
            if p != -1 {
                children[p as usize].push(i);
            }
        }
        let mut ans = 0;
        Solution::rc(&children, &chars, 0, &mut ans);
        ans
    }
}