题目
题目链接：
https://www.nowcoder.com/practice/a77b4f3d84bf4a7891519ffee9376df3
思路
核心就是树的最大直径(globalMax)一定是以某一个node为root最长的两个path-to-leaf.
就是普通dfs的同时算路径长度。
时间: O(n), DFS一次
空间: O(n)
参考答案Java
import java.util.*;
/*
* public class Interval {
* int start;
* int end;
* public Interval(int start, int end) {
* this.start = start;
* this.end = end;
* }
* }
*/
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 树的直径
* @param n int整型 树的节点个数
* @param Tree_edge Interval类一维数组 树的边
* @param Edge_value int整型一维数组 边的权值
* @return int整型
*/
public int solve (int n, Interval[] Tree_edge, int[] Edge_value) {
//核心就是树的最大直径(globalMax)一定是以某一个node为root最长的两个path-to-leaf.
//就是普通dfs的同时算路径长度。
//
//时间: O(n), DFS一次
//空间: O(n)
// node_id -> { {nei1, cost1}, {nei2, cost2}, ... }
Map<Integer, List<int[]>> graph = new HashMap<>();
int[] globalMax = {0};
// construct graph
for (int i = 0; i < n ; i++) {
graph.put(i, new ArrayList<int[]>());
}
for (int i = 0; i < n - 1 ; i++) {
// treat tree edge as bi-directional
int from = Tree_edge[i].start;
int to = Tree_edge[i].end;
graph.get(from).add(new int[] {to, Edge_value[i]});
graph.get(to).add(new int[] {from, Edge_value[i]});
}
// 因为edge是bi-directional, 随便从哪个node开始dfs都一样。这里用了node-0.
// -1 作为parent
dfs(graph, globalMax, 0, -1);
return globalMax[0];
}
// returns the max cost path-to-leaf from this root.
public int dfs(Map<Integer, List<int[]>> graph, int[] ans, int root,
int parent) {
int maxCost = 0;
int maxCost2 = 0;
for (int[] nexts : graph.get(root)) {
// NOTE: BaseCase (i.e. leaf) has only 1 nei, which is the parent
// thus leaf will return maxCost = 0 directly.
if (nexts[0] == parent) continue;
// recursively finds the max cost path to any leaf
int cost = dfs(graph, ans, nexts[0], root) + nexts[1];
// keep 2 largest cost
if (cost >= maxCost) {
maxCost2 = maxCost;
maxCost = cost;
} else if(cost >=maxCost2){
maxCost2 = cost;
}
}
// check if the 2 largest path is the global max
int curcost = maxCost + maxCost2;
if (curcost > ans[0]) {
ans[0] = curcost;
}
return maxCost;
}
}
参考答案Go
package main
import . "nc_tools"
/*
* type Interval struct {
* Start int
* End int
* }
*/
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 树的直径
* @param n int整型 树的节点个数
* @param Tree_edge Interval类一维数组 树的边
* @param Edge_value int整型一维数组 边的权值
* @return int整型
*/
func solve(n int, Tree_edge []*Interval, Edge_value []int) int {
//核心就是树的最大直径(globalMax)一定是以某一个node为root最长的两个path-to-leaf.
//就是普通dfs的同时算路径长度。
//
//时间: O(n), DFS一次
//空间: O(n)
// node_id -> { {nei1, cost1}, {nei2, cost2}, ... }
graph := map[int][]*Node{} //map表示图
// construct graph
for i := 0; i < n; i++ {
graph[i] = []*Node{}
}
for i := 0; i < n-1; i++ {
// treat tree edge as bi-directional
from := Tree_edge[i].Start
to := Tree_edge[i].End
graph[from] = append(graph[from], &Node{to, Edge_value[i]})
graph[to] = append(graph[to], &Node{from, Edge_value[i]})
}
// 因为edge是bi-directional, 随便从哪个node开始dfs都一样。这里用了node-0.
// -1 作为parent
ans := []int{0}
dfs(graph, &ans, 0, -1)
return ans[0]
}
func dfs(graph map[int][]*Node, ans *[]int, root, parent int) int {
maxCost := 0
maxCost2 := 0
for _, nexts := range graph[root] {
// NOTE: BaseCase (i.e. leaf) has only 1 nei, which is the parent
// thus leaf will return maxCost = 0 directly.
if nexts.to == parent {
continue
}
// recursively finds the max cost path to any leaf
cost := dfs(graph, ans, nexts.to, root) + nexts.cost
// keep 2 largest cost
if cost >= maxCost {
maxCost2 = maxCost
maxCost = cost
} else if cost >= maxCost2 {
maxCost2 = cost
}
}
// check if the 2 largest path is the global max
cursot := maxCost + maxCost2
if cursot > (*ans)[0] {
(*ans)[0] = cursot
}
return maxCost
}
type Node struct {
to int
cost int
}
参考答案PHP
<?php
/*class Interval{
var $start = 0;
var $end = 0;
function __construct($a, $b){
$this->start = $a;
$this->end = $b;
}
}*/
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 树的直径
* @param n int整型 树的节点个数
* @param Tree_edge Interval类一维数组 树的边
* @param Edge_value int整型一维数组 边的权值
* @return int整型
*/
function solve( $n , $Tree_edge , $Edge_value )
{
//核心就是树的最大直径(globalMax)一定是以某一个node为root最长的两个path-to-leaf.
//就是普通dfs的同时算路径长度。
//
//时间: O(n), DFS一次
//空间: O(n)
// node_id -> { {nei1, cost1}, {nei2, cost2}, ... }
$graph = [];
for($i=0;$i<$n;$i++){
$graph[$i] = [];
}
// construct graph
for($i=0;$i<$n-1;$i++){
$from = $Tree_edge[$i]->start;
$to = $Tree_edge[$i]->end;
$graph[$from][count($graph[$from])] = [$to,$Edge_value[$i]];
$graph[$to][count($graph[$to])] = [$from,$Edge_value[$i]];
}
$ans = [0];
// 因为edge是bi-directional, 随便从哪个node开始dfs都一样。这里用了node-0.
// -1 作为parent
dfs($graph,$ans,0,-1);
return $ans[0];
}
// returns the max cost path-to-leaf from this root.
function dfs($graph,&$ans,$root,$parent){
$max1 =0;
$max2 = 0;
foreach ($graph[$root] as $nexts) {
// NOTE: BaseCase (i.e. leaf) has only 1 nei, which is the parent
// thus leaf will return maxCost = 0 directly.
if($nexts[0] == $parent) continue; //不走回头路
// recursively finds the max cost path to any leaf
$cost = dfs($graph,$ans,$nexts[0],$root)+$nexts[1];
// keep 2 largest cost
if($cost >= $max1){
$max2=$max1;
$max1 = $cost;
}else if($cost >=$max2){
$max2 = $cost;
}
}
// check if the 2 largest path is the global max
$curcost = $max1+$max2;
if($curcost > $ans[0]){
$ans[0] = $curcost;
}
return $max1;
}
