演算法練習week5--leetcode684
一個無向圖是否存在環
演算法1:
我們知道對於環 1-2-3-4-1,每個節點的度都是2,基於此我們有如下演算法(這是類似於有向圖的拓撲排序):
- 求出圖中所有頂點的度
- 刪除圖中所有度 <=1 的頂點以及與該頂點相關的邊,把與這些邊相關的頂點的度減一
- 如果還有度<=1的頂點重複步驟2
- 最後如果還存在未被刪除的頂點,則表示有環;否則沒有環
時間複雜度為O(E+V)
,其中E、V分別為圖中邊和頂點的數目。
演算法2:
深度優先遍歷該圖,如果在遍歷的過程中,發現某個節點有一條邊指向已經訪問過的節點,並且這個已訪問過的節點不是當前節點的父節點(這裡的父節點表示dfs遍歷順序中的父節點),則表示存在環。但是我們不能僅僅使用一個bool陣列來標誌節點是否訪問過。
對每個節點分為三種狀態,白、灰、黑。
開始時所有節點都是白色,當開始訪問某個節點時該節點變為灰色,當該節點的所有鄰接點都訪問完,該節點顏色變為黑色。
那麼我們的演算法則為:如果遍歷的過程中發現某個節點有一條邊指向顏色為灰的節點,那麼存在環。
LeeCode題目
LeetCode題目:684. Redundant Connection
In this problem, a tree is an undirected 無向圖 graph that is connected and has no cycles.
The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.
Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.
Example 1:
Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
Example 1
Example 2:
Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
Example 2
Note:
- The size of the input 2D-array will be between 3 and 1000.
- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
//int eSize = edges.size();
vector<int> start(3000);
//memset(start, 0, sizeof(int)*eSize);
for (int i = 0; i < 3000; i++) {
start[i] = i;
}
int tStart, tEnd;
for (int i = 0; i < edges.size(); i++) {
tStart = edges[i][0];
tEnd = edges[i][1];
if (start[tStart] == start[tEnd]) {
return edges[i];
}
int needUpdate = start[tEnd];
for (int j = 0; j < 3000; j++) {
if (start[j] == needUpdate) {
start[j] = start[tStart];
}
}
}
}
};
Note:上述演算法的時間複雜度是O(n^2)