Code & Func

第21天。

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.

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Example 1:
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Input: [[1,3], [0,2], [1,3], [0,2]]
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Output: true
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Explanation:
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The graph looks like this:
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0----1
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|    |
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|    |
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3----2
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We can divide the vertices into two groups: {0, 2} and {1, 3}.
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Example 2:
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Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
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Output: false
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Explanation:
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The graph looks like this:
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0----1
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| \  |
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|  \ |
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3----2
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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].

这是一道关于图的问题，题目的意思很简单，就是要判断一个图是不是一个二部图，所谓的二部图，就是一个图可以把所有节点划分到两个不相交的两个集合，这两个集合内部没有边相连。

我们可以对图进行一次遍历，遍历的时候对节点进行着色，着色的规律是这样的，当从一个节点跳到另一个节点的时候，我们就切换一次颜色（共有三种颜色，其中一种表示没有访问，即白色）。因为遍历完了之后，整个图的节点就被划分成两部分了，接下来我们只需要判断所有节点的邻居是否和它是不同色的即可。代码如下：

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char color;
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bool isBipartite(vector<vector<int>>& graph) {
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    int size = graph.size();
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    vector<char> flags(size, 'w');
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    // w g b
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    color = 'b';
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    for(int i = 0;i < size;i++) {
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        if (flags[i] == 'w') {
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            dfs(graph, flags, i);
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        }
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    }
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    for(int i = 0;i < size;i++) {
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        for(auto j: graph[i]) {
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            if (flags[i] == flags[j]) return false;
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        }
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    }
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    return true;
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}
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void dfs(vector<vector<int>> &graph, vector<char> &flags, int index) {
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    flags[index] = color;
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    for(auto j: graph[index]) {
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        if (flags[j] == 'w') {
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            color = ~color;
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            dfs(graph, flags, j);
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            color = ~color;
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        }
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    }
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}