Given an undirected graph
, return true
if and only if it is bipartite.
Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.
The graph is given in the following form: graph[i]
is a list of indexes j
for which the edge between nodes i
and j
exists. Each node is an integer between 0
and graph.length - 1
. There are no self edges or parallel edges: graph[i]
does not contain i
, and it doesn't contain any element twice.
Example 1: Input: [[1,3], [0,2], [1,3], [0,2]] Output: true Explanation: The graph looks like this: 0----1 | | | | 3----2 We can divide the vertices into two groups: {0, 2} and {1, 3}.Example 2: Input: [[1,2,3], [0,2], [0,1,3], [0,2]] Output: false Explanation: The graph looks like this: 0----1 | \ | | \ | 3----2 We cannot find a way to divide the set of nodes into two independent subsets.
Note:
graph
will have length in range[1, 100]
.graph[i]
will contain integers in range[0, graph.length - 1]
.graph[i]
will not containi
or duplicate values.- The graph is undirected: if any element
j
is ingraph[i]
, theni
will be ingraph[j]
.
分析
这道题需要利用广搜的办法来解决。为了保证将所有的Node分成两个集合,并且每条边的两个node都在两个集合中,可以首先确定一个node例如0在part1中,那么0的所有边的另外一端必然在part2中。我们可以使用队列进行BFS。当遇到没有边的node时,可以随便放进一个集合,也可以不放。也要考虑一种情况就是一个graph可能会有两个独立的图。如下所示
0----1 4----5 | | | | | | | | 3----2 6----7
所以我们只把0塞进去做广搜没法遍历所有的node,那么需要维护一个没有访问过的node的set,每次队列为空后,就从set中拿出一个node塞进queue进去搜索。
Code
class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
set<int> p1;
set<int> p2;
int len = graph.size();
if (len == 0)
return false;
queue<pair<int, int>> q;
set<int> v;
for (int i = 0; i < len; i ++)
{
v.insert(i);
}
while (v.size() != 0)
{
q.push(make_pair(*v.begin(), 0));
v.erase(v.begin());
while (!q.empty())
{
int index = q.front().first;
int part = q.front().second;
q.pop();
v.erase(index);
if (part == 0)
{
if (p2.find(index) != p2.end())
return false;
if (p1.find(index) != p1.end())
continue;
p1.insert(index);
}
else
{
if (p1.find(index) != p1.end())
return false;
if (p2.find(index) != p2.end())
continue;
p2.insert(index);
}
for (int i = 0; i < graph[index].size(); i ++)
{
q.push(make_pair(graph[index][i], 1-part));
}
}
}
return true;
}
};
运行效率
Runtime: 32 ms, faster than 31.39% of C++ online submissions for Is Graph Bipartite?.
Memory Usage: 13.1 MB, less than 7.03% of C++ online submissions for Is Graph Bipartite?.