找到与 X 绝对差最大的节点
原文:https://www . geesforgeks . org/find-the-node-with-x-的绝对差值给出最大值/
给定一棵树，所有节点的权重和一个整数 x ，任务是找到一个节点 i ，使得|权重【I】–x |最大。 举例:
输入:
x = 15 输出: 1 节点 1:| 5–15 | = 10 节点 2:| 10–15 | = 5 节点 3: |11 -15| = 4 节点 4:| 8–15 | = 7 节点 5: |6 -15| = 9
方法:在树上执行 dfs 并跟踪其与的加权绝对差 x 给出最大值的节点。 以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int maximum = INT_MIN, x, ans;
vector<int> graph[100];
vector<int> weight(100);
// Function to perform dfs to find
// the maximum value
void dfs(int node, int parent)
{
// If current value is more than
// the current maximum
if (maximum < abs(weight[node] - x)) {
maximum = abs(weight[node] - x);
ans = node;
}
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
int main()
{
x = 15;
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
// Edges of the tree
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
dfs(1, 1);
cout << ans;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
static int maximum = Integer.MIN_VALUE, x, ans;
static Vector<Vector<Integer>> graph=new Vector<Vector<Integer>>();
static Vector<Integer> weight=new Vector<Integer>();
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
// If current value is more than
// the current maximum
if (maximum < Math.abs(weight.get(node) - x))
{
maximum = Math.abs(weight.get(node) - x);
ans = node;
}
for (int i = 0; i < graph.get(node).size(); i++)
{
if (graph.get(node).get(i) == parent)
continue;
dfs(graph.get(node).get(i), node);
}
}
// Driver code
public static void main(String args[])
{
x = 15;
// Weights of the node
weight.add(0);
weight.add(5);
weight.add(10);;
weight.add(11);;
weight.add(8);
weight.add(6);
for(int i = 0; i < 100; i++)
graph.add(new Vector<Integer>());
// Edges of the tree
graph.get(1).add(2);
graph.get(2).add(3);
graph.get(2).add(4);
graph.get(1).add(5);
dfs(1, 1);
System.out.println( ans);
}
}
// This code is contributed by Arnab Kundu
Python 3
# Python implementation of the approach
from sys import maxsize
# Function to perform dfs to find
# the minimum value
def dfs(node, parent):
global minimum, graph, weight, x, ans
# If current value is less than
# the current minimum
if minimum < abs(weight[node] - x):
minimum = abs(weight[node] - x)
ans = node
for to in graph[node]:
if to == parent:
continue
dfs(to, node)
# Driver Code
if __name__ == "__main__":
minimum = -maxsize
graph = [[] for i in range(100)]
weight = [0] * 100
x = 15
ans = 0
# Weights of the node
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
dfs(1, 1)
print(ans)
# This code is contributed by
# sanjeev2552
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int maximum = int.MinValue, x, ans;
static List<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
// If current value is more than
// the current maximum
if (maximum < Math.Abs(weight[node] - x))
{
maximum = Math.Abs(weight[node] - x);
ans = node;
}
for (int i = 0; i < graph[node].Count; i++)
{
if (graph[node][i] == parent)
continue;
dfs(graph[node][i], node);
}
}
// Driver code
public static void Main(String []args)
{
x = 15;
// Weights of the node
weight.Add(0);
weight.Add(5);
weight.Add(10);;
weight.Add(11);;
weight.Add(8);
weight.Add(6);
for(int i = 0; i < 100; i++)
graph.Add(new List<int>());
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
Console.WriteLine( ans);
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript implementation of the approach
let maximum = Number.MIN_VALUE, x, ans;
let graph= [];
let weight=[];
// Function to perform dfs to find
// the maximum value
function dfs(node,parent)
{
// If current value is more than
// the current maximum
if (maximum < Math.abs(weight[node] - x))
{
maximum = Math.abs(weight[node] - x);
ans = node;
}
for (let i = 0; i < graph[node].length; i++)
{
if (graph[node][i] == parent)
continue;
dfs(graph[node][i], node);
}
}
// Driver code
x = 15;
// Weights of the node
weight.push(0);
weight.push(5);
weight.push(10);;
weight.push(11);;
weight.push(8);
weight.push(6);
for(let i = 0; i < 100; i++)
graph.push([]);
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write( ans);
// This code is contributed by unknown2108
</script>
Output:
1
复杂度分析:
- 时间复杂度: O(N)。 在 dfs 中，树的每个节点都被处理一次，因此如果树中总共有 N 个节点，由于 dfs 而导致的复杂性是 O(N)。因此，时间复杂度为 O(N)。
- 辅助空间: O(1)。 不需要任何额外的空间，所以空间复杂度不变。
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