LeetCode1466. Reorder Routes to Make All Paths Lead to the City Zero

[Med] LeetCode 1466. Reorder Routes to Make All Paths Lead to the City Zero

连接: https://leetcode.com/problems/reorder-routes-to-make-all-paths-lead-to-the-city-zero/

题目描述：
There are n cities numbered from 0 to n-1 and n-1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.

Roads are represented by connections where connections[i] = [a, b] represents a road from city a to b.

This year, there will be a big event in the capital (city 0), and many people want to travel to this city.

Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.

It’s guaranteed that each city can reach the city 0 after reorder.

Example 1:

Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
Output: 3
Explanation: Change the direction of edges show in red such that each node can reach the node 0

Example 2:

Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
Output: 2
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 3:

Input: n = 3, connections = [[1,0],[2,0]]
Output: 0

Constraints:

• 2 <= n <= 5 * 10^4
• connections.length == n-1
• connections[i].length == 2
• 0 <= connections[i][0], connections[i][1] <= n-1
• connections[i][0] != connections[i][1]

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Tag: Tree, DFS
解题思路
题目告诉咱们有n个城市，每两个城市之间只有一条道路相连，而且这一条道路是一条有方向的道路。如今咱们要让全部的城市均可以到达城市0 (到达指的是能够通过0个或者多个有向道路到达)。问，咱们须要将多少条城市道路的方向调转过来才能够达到这样的目的。题目当中能够保证结果能够出现。

由于题目中说，咱们保证两个城市之间只有一条路能够到达(这一条路能够通过多个城市)。也就是说这里实际上是一个有向无环的树。咱们要作的就是从这棵树的root (也就是城市0)出发，通过全部的子节点。若是在这个过程中父节点到子节点的路径和给出的connections路径相反的话，那么咱们就须要转换一次方向使得路径是从父节点往子节点方向去的。每一次转换咱们都将转换的总次数加一，这样咱们最后就会构成一个从root 为0的树，方向为父节点指向子节点。由于咱们最后想要的结果是全部的子节点指向父节点的一棵树，因此咱们最后还要用edge的总和，也就是n-1, 减去咱们累加出来的结果。才是将全部节点指向root的须要转换的次数。

为了构成这棵树，咱们首先构建一个无向图。这个无向图咱们使用hashmap来构造。两个点之间互相有所链接。同时为了在遍历这个无向图的时候快速定位这个方向和connections里面的方向是否是对的。咱们再使用一个hashset，里面存上城市原道路的方向 。这样在遍历咱们无向图的时候就能够快速的定位了。同时为了不造成环，咱们再使用一个长度为n的数组记录咱们已经遍历过的节点。

作完这些以后咱们从0节点出发。每一次咱们都从map当中获取从source节点能够到达的全部children节点。若是这个节点已经被visit过了，则跳过。不然若是当前咱们遍历的两个节点的方向没有出如今direction hashset中，则增长total的值。

最后是返回n-1-total的值

TIME & SPACE: O(n)
解法一:

class Solution {
    int total =0;
    public int minReorder(int n, int[][] connections) {
        //注意不能添加新的edge
        Map<Integer, List<Integer>>  map = new HashMap<>();
        Set<String> direction = new HashSet<>();
        
        for(int[] connection : connections){
            int a = connection[0], b = connection[1];
            direction.add(a+"-"+b);

            if(!map.containsKey(b)) map.put(b, new ArrayList());
            if(!map.containsKey(a)) map.put(a, new ArrayList());
            map.get(a).add(b);
            map.get(b).add(a);
        }
        
        dfs(0, map, direction, new int[n]);
        return n-1-total;
    }
    
    public void dfs(int source, Map<Integer, List<Integer>> map, Set<String> direction, int[] visited){
        if(visited[source]==1) return;
        visited[source] = 1;
        
        for(int child : map.get(source)){
            if(visited[child]==1) continue;
            if(!direction.contains(source+"-"+child)){
                total++;
            }
            dfs(child, map, direction, visited);
        }
    }
}