785. Is Graph Bipartite? (M)
https://leetcode.com/problems/is-graph-bipartite/
There is an undirected graph with n
nodes, where each node is numbered between 0
and n - 1
. You are given a 2D array graph
, where graph[u]
is an array of nodes that node u
is adjacent to. More formally, for each v
in graph[u]
, there is an undirected edge between node u
and node v
. The graph has the following properties:
There are no self-edges (
graph[u]
does not containu
).There are no parallel edges (
graph[u]
does not contain duplicate values).If
v
is ingraph[u]
, thenu
is ingraph[v]
(the graph is undirected).The graph may not be connected, meaning there may be two nodes
u
andv
such that there is no path between them.
A graph is bipartite if the nodes can be partitioned into two independent sets A
and B
such that every edge in the graph connects a node in set A
and a node in set B
.
Return true
if and only if it is bipartite.
Example 1:
Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.
Example 2:
Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.
Constraints:
graph.length == n
1 <= n <= 100
0 <= graph[u].length < n
0 <= graph[u][i] <= n - 1
graph[u]
does not containu
.All the values of
graph[u]
are unique.If
graph[u]
containsv
, thengraph[v]
containsu
.
Solution:
比如题目给的例子,输入的邻接表 graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
,也就是这样一幅图:
显然无法对节点着色使得每两个相邻节点的颜色都不相同,所以算法返回 false。
但如果输入 graph = [[1,3],[0,2],[1,3],[0,2]]
,也就是这样一幅图:
如果把节点 {0, 2}
涂一个颜色,节点 {1, 3}
涂另一个颜色,就可以解决「双色问题」,所以这是一幅二分图,算法返回 true。
结合之前的代码框架,我们可以额外使用一个 color
数组来记录每个节点的颜色,从而写出解法代码:
// 记录图是否符合二分图性质
private boolean ok = true;
// 记录图中节点的颜色,false 和 true 代表两种不同颜色
private boolean[] color;
// 记录图中节点是否被访问过
private boolean[] visited;
// 主函数,输入邻接表,判断是否是二分图
public boolean isBipartite(int[][] graph) {
int n = graph.length;
color = new boolean[n];
visited = new boolean[n];
// 因为图不一定是联通的,可能存在多个子图
// 所以要把每个节点都作为起点进行一次遍历
// 如果发现任何一个子图不是二分图,整幅图都不算二分图
for (int v = 0; v < n; v++) {
if (!visited[v]) {
traverse(graph, v);
}
}
return ok;
}
// DFS 遍历框架
private void traverse(int[][] graph, int v) {
// 如果已经确定不是二分图了,就不用浪费时间再递归遍历了
if (!ok) return;
visited[v] = true;
for (int w : graph[v]) {
if (!visited[w]) {
// 相邻节点 w 没有被访问过
// 那么应该给节点 w 涂上和节点 v 不同的颜色
color[w] = !color[v];
// 继续遍历 w
traverse(graph, w);
} else {
// 相邻节点 w 已经被访问过
// 根据 v 和 w 的颜色判断是否是二分图
if (color[w] == color[v]) {
// 若相同,则此图不是二分图
ok = false;
}
}
}
}
这就是解决「双色问题」的代码,如果能成功对整幅图染色,则说明这是一幅二分图,否则就不是二分图。
接下来看一下 BFS 算法的逻辑:
// 记录图是否符合二分图性质
private boolean ok = true;
// 记录图中节点的颜色,false 和 true 代表两种不同颜色
private boolean[] color;
// 记录图中节点是否被访问过
private boolean[] visited;
public boolean isBipartite(int[][] graph) {
int n = graph.length;
color = new boolean[n];
visited = new boolean[n];
for (int v = 0; v < n; v++) {
if (!visited[v]) {
// 改为使用 BFS 函数
bfs(graph, v);
}
}
return ok;
}
// 从 start 节点开始进行 BFS 遍历
private void bfs(int[][] graph, int start) {
Queue<Integer> q = new LinkedList<>();
visited[start] = true;
q.offer(start);
while (!q.isEmpty() && ok) {
int v = q.poll();
// 从节点 v 向所有相邻节点扩散
for (int w : graph[v]) {
if (!visited[w]) {
// 相邻节点 w 没有被访问过
// 那么应该给节点 w 涂上和节点 v 不同的颜色
color[w] = !color[v];
// 标记 w 节点,并放入队列
visited[w] = true;
q.offer(w);
} else {
// 相邻节点 w 已经被访问过
// 根据 v 和 w 的颜色判断是否是二分图
if (color[w] == color[v]) {
// 若相同,则此图不是二分图
ok = false;
}
}
}
}
}
核心逻辑和刚才实现的 traverse
函数(DFS 算法)完全一样,也是根据相邻节点 v
和 w
的颜色来进行判断的。关于 BFS 算法框架的探讨,详见前文 BFS 算法框架 和 Dijkstra 算法模板,这里就不展开了。
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