1615. 最大网络秩

题目：

思路：

【1】计数枚举的方式

【2】贪心的方式，只选取出拥有边最多的节点和次多的节点进行计算

代码展示：

【1】计数枚举的方式

//时间18 ms 击败 26.27%
//内存42.8 MB 击败 80.51%
class Solution {
    public int maximalNetworkRank(int n, int[][] roads) {
         HashMap<Integer, HashSet<Integer>> map = new HashMap<>();
        for (int i = 0; i < n; i++){
            map.put(i,new HashSet<Integer>());
        }
        for (int[] path : roads){
            HashSet<Integer> tem1 = map.get(path[0]);
            tem1.add(path[1]);
            HashSet<Integer> tem2 = map.get(path[1]);
            tem2.add(path[0]);
        }
        
        // 进行枚举的方式，复杂度一下子到了O(N^2)
        int p = 0;
        for (int i = 0; i < n - 1; i++){
            HashSet<Integer> tem1 = map.get(i);
            for (int j = i+1; j < n; j++){
                HashSet<Integer> tem2 = map.get(j);
                if (tem1.contains(j)){
                    p = Math.max(p,tem1.size()+tem2.size()-1);
                }else {
                    p = Math.max(p,tem1.size()+tem2.size());
                }

            }
        }
        return p;
    }
}

【2】贪心的方式，只选取出拥有边最多的节点和次多的节点进行计算（相当于在第一种的方法上进行了截枝）

//时间1 ms 击败 100%
//内存43.5 MB 击败 11.86%
class Solution {
    public int maximalNetworkRank(int n, int[][] roads) {
        int[] degrees = new int[n];
        for (int[] road : roads) {
            ++degrees[road[0]];
            ++degrees[road[1]];
        }

        int max1 = 0, max2 = 0;
        for (int degree : degrees) {
            if (degree > max1) {
                max2 = max1;
                max1 = degree;
            } else if (degree > max2 && degree < max1) {
                max2 = degree;
            }
        }

        int count1 = 0, count2 = 0;
        for (int degree : degrees) {
            if (degree == max1) {
                ++count1;
            }
            if (degree == max2) {
                ++count2;
            }
        }

        if (count1 == 1) {
            int edgeCount = 0;
            for (int[] road : roads) {
                if (degrees[road[0]] == max1 && degrees[road[1]] == max2 || degrees[road[0]] == max2 && degrees[road[1]] == max1) {
                    ++edgeCount;
                }
            }
            return edgeCount == count2 ? max1 + max2 - 1 : max1 + max2;
        } else {
            int edgeCount = 0, limit = count1 * (count1 - 1) / 2;
            for (int[] road : roads) {
                if (degrees[road[0]] == max1 && degrees[road[1]] == max1) {
                    ++edgeCount;
                }
            }
            return edgeCount == limit ? max1 * 2 - 1 : max1 * 2;
        }
    }
}