785. Is Graph Bipartite?

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 contain `i` or duplicate values.
• The graph is undirected: if any element `j` is in `graph[i]`, then `i` will be in `graph[j]`.

``````class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
vector<int> colors(graph.size());
for (int i = 0; i < graph.size(); ++i) {
if (colors[i] == 0 && !valid(graph, 1, i, colors)) {
return false;
}
}
return true;
}
bool valid(vector<vector<int>>& graph, int color, int cur, vector<int>& colors) {
if (colors[cur] != 0) return colors[cur] == color;
colors[cur] = color;
for (int i : graph[cur]) {
if (!valid(graph, -1 * color, i, colors)) {
return false;
}
}
return true;
}
};
``````

``````class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
vector<int> colors(graph.size());
for (int i = 0; i < graph.size(); ++i) {
if (colors[i] != 0) continue;
colors[i] = 1;
queue<int> q{{i}};
while (!q.empty()) {
int t = q.front(); q.pop();
for (auto a : graph[t]) {
if (colors[a] == colors[t]) return false;
if (colors[a] == 0) {
colors[a] = -1 * colors[t];
q.push(a);
}
}
}
}
return true;
}
};
``````

``````class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
vector<int> root(graph.size());
for (int i = 0; i < graph.size(); ++i) root[i] = i;
for (int i = 0; i < graph.size(); ++i) {
if (graph[i].empty()) continue;
int x = find(root, i), y = find(root, graph[i][0]);
if (x == y) return false;
for (int j = 1; j < graph[i].size(); ++j) {
int parent = find(root, graph[i][j]);
if (x == parent) return false;
root[parent] = y;
}
}
return true;
}
int find(vector<int>& root, int i) {
return root[i] == i ? i : find(root, root[i]);
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/785

Possible Bipartition

https://leetcode.com/problems/is-graph-bipartite/

https://leetcode.com/problems/is-graph-bipartite/discuss/115487/Java-Clean-DFS-solution-with-Explanation