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[LeetCode] 2374. Node With Highest Edge Score

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You are given a directed graph with n nodes labeled from 0 to n - 1, where each node has exactly one outgoing edge.

The graph is represented by a given 0-indexed integer array edges of length n, where edges[i] indicates that there is a directed edge from node i to node edges[i].

The edge score of a node i is defined as the sum of the labels of all the nodes that have an edge pointing to i.

Return the node with the highest edge score. If multiple nodes have the same edge score, return the node with the smallest index.

Example 1:

Input: edges = [1,0,0,0,0,7,7,5]
Output: 7
Explanation:
- The nodes 1, 2, 3 and 4 have an edge pointing to node 0. The edge score of node 0 is 1 + 2 + 3 + 4 = 10.
- The node 0 has an edge pointing to node 1. The edge score of node 1 is 0.
- The node 7 has an edge pointing to node 5. The edge score of node 5 is 7.
- The nodes 5 and 6 have an edge pointing to node 7. The edge score of node 7 is 5 + 6 = 11.
Node 7 has the highest edge score so return 7.

Example 2:

Input: edges = [2,0,0,2]
Output: 0
Explanation:
- The nodes 1 and 2 have an edge pointing to node 0. The edge score of node 0 is 1 + 2 = 3.
- The nodes 0 and 3 have an edge pointing to node 2. The edge score of node 2 is 0 + 3 = 3.
Nodes 0 and 2 both have an edge score of 3. Since node 0 has a smaller index, we return 0.

Constraints:

  • n == edges.length
  • 2 <= n <= 105
  • 0 <= edges[i] < n
  • edges[i] != i

边积分最高的节点。

给你一个有向图,图中有 n 个节点,节点编号从 0 到 n - 1 ,其中每个节点都 恰有一条 出边。

图由一个下标从 0 开始、长度为 n 的整数数组 edges 表示,其中 edges[i] 表示存在一条从节点 i 到节点 edges[i] 的 有向 边。

节点 i 的 边积分 定义为:所有存在一条指向节点 i 的边的节点的 编号 总和。

返回 边积分 最高的节点。如果多个节点的 边积分 相同,返回编号 最小 的那个。

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/node-with-highest-edge-score
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

题目以数组的形式给了一些点的联通关系,但是这道题跟图论毫无关系,就是一道数组题。

我们创建一个和 edges 等长的数组记录每个 node 的得分。对于 input 数组,下标 i 代表的是一条有向边的起点,i 位置上的值代表的是这条有向边的终点。对于每个节点,他的边积分等于能到达此节点的所有的边的起点的和。比如第一个例子,有多条边指向 0,那么 0 的边积分就等于所有指向 0 的节点的 index 的和。这里我展示一下分别用hashmap和数组实现的代码,速度差距还是挺大的。注意由于节点的个数会大到 10^5,所以记录每个节点的变积分需要用 long 型。

时间O(n)

空间O(n)

hashmap实现

 1 class Solution {
 2     public int edgeScore(int[] edges) {
 3         HashMap<Integer, Long> map = new HashMap<>();
 4         int len = edges.length;
 5         for (int i = 0; i < edges.length; i++) {
 6             int from = i;
 7             int to = edges[i];
 8             if (!map.containsKey(to)) {
 9                 map.put(to, 0L);
10             }
11             long sum = map.getOrDefault(to, 0L);
12             map.put(to, sum + from);
13         }
14         
15         long max = 0;
16         int res = 0;
17         for (int i = 0; i < len; i++) {
18             if (map.containsKey(i)) {
19                 long val = map.get(i);
20                 if (val > max) {
21                     res = i;
22                     max = val;
23                 }
24             }
25         }
26         return res;
27     }
28 }

 

数组实现

 1 class Solution {
 2     public int edgeScore(int[] edges) {
 3         long[] map = new long[edges.length];
 4         for (int i = 0; i < edges.length; i++) {
 5             int from = i;
 6             int to = edges[i];
 7             map[to] += from;
 8         }
 9 
10         int res = 0;
11         long max = 0L;
12         for (int i = 0; i < map.length; i++) {
13             if (map[i] > max) {
14                 max = map[i];
15                 res = i;
16             }
17         }
18         return res;
19     }
20 }

 

LeetCode 题目总结

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