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]

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].

這道題博主在最開始做的時候，看了半天，楞是沒弄懂輸出數據的意思，博主開始以為給的是邊，後來發現跟圖對應不上，就懵逼了，後來是通過研究論壇上大神們的解法，才總算搞懂了題目的意思，原來輸入數組中的graph[i]，表示頂點i所有相鄰的頂點，比如對於例子1來說，頂點0和頂點1，3相連，頂點1和頂點0，2相連，頂點2和結點1，3相連，頂點3和頂點0，2相連。這道題讓我們驗證給定的圖是否是二分圖，所謂二分圖，就是可以將圖中的所有頂點分成兩個不相交的集合，使得同一個集合的頂點不相連。為了驗證是否有這樣的兩個不相交的集合存在，我們采用一種很機智的染色法，大體上的思路是要將相連的兩個頂點染成不同的顏色，一旦在染的過程中發現有兩連的兩個頂點已經被染成相同的顏色，說明不是二分圖。這裏我們使用兩種顏色，分別用1和-1來表示，初始時每個頂點用0表示未染色，然後遍歷

解法一：

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;
    }
};

解法二：

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) colors[i] = 1;
            for (auto a : graph[i]) {
                if (colors[a] == colors[i]) return false;
                colrs[a] = -1 * colors[a];
            }
        }
        return true;
    }
};

參考資料：

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

https://leetcode.com/problems/is-graph-bipartite/discuss/115723/C++-short-iterative-solution-with-comments

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[LeetCode] Is Graph Bipartite? 是二分圖麽？