Add solution 0785

leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite.go

@@ -0,0 +1,35 @@
+package leetcode
+
+// DFS 染色，1 是红色，0 是绿色，-1 是未染色
+func isBipartite(graph [][]int) bool {
+	colors := make([]int, len(graph))
+	for i := range colors {
+		colors[i] = -1
+	}
+	for i := range graph {
+		if !dfs(i, graph, colors, -1) {
+			return false
+		}
+	}
+	return true
+}
+
+func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
+	if colors[n] == -1 {
+		if parentCol == 1 {
+			colors[n] = 0
+		} else {
+			colors[n] = 1
+		}
+	} else if colors[n] == parentCol {
+		return false
+	} else if colors[n] != parentCol {
+		return true
+	}
+	for _, c := range graph[n] {
+		if !dfs(c, graph, colors, colors[n]) {
+			return false
+		}
+	}
+	return true
+}

leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question785 struct {
+	para785
+	ans785
+}
+
+// para 是参数
+// one 代表第一个参数
+type para785 struct {
+	graph [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans785 struct {
+	one bool
+}
+
+func Test_Problem785(t *testing.T) {
+
+	qs := []question785{
+
+		{
+			para785{[][]int{{1, 3}, {0, 2}, {1, 3}, {0, 2}}},
+			ans785{true},
+		},
+
+		{
+			para785{[][]int{{1, 2, 3}, {0, 2}, {0, 1, 3}, {0, 2}}},
+			ans785{false},
+		},
+
+		{
+			para785{[][]int{{1, 2, 3}, {0, 2, 3}, {0, 1, 3}, {0, 1, 2}}},
+			ans785{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 785------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans785, q.para785
+		fmt.Printf("【input】:%v       【output】:%v\n", p, isBipartite(p.graph))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0785.Is-Graph-Bipartite/README.md

@@ -0,0 +1,98 @@
+# [785. Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite/)
+
+
+## 题目
+
+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]`.
+
+## 题目大意
+
+给定一个无向图 graph，当这个图为二分图时返回 true。
+
+graph 将会以邻接表方式给出，graph[i] 表示图中与节点i相连的所有节点。每个节点都是一个在 0 到 graph.length-1 之间的整数。这图中没有自环和平行边： graph[i] 中不存在 i，并且 graph[i] 中没有重复的值。
+
+注意:
+
+- graph 的长度范围为 [1, 100]。
+- graph[i] 中的元素的范围为 [0, graph.length - 1]。
+- graph[i] 不会包含 i 或者有重复的值。
+- 图是无向的: 如果 j 在 graph[i] 里边, 那么 i 也会在 graph[j] 里边。
+
+## 解题思路
+
+- 判断一个无向图是否是二分图。二分图的定义：如果我们能将一个图的节点集合分割成两个独立的子集 A 和 B，并使图中的每一条边的两个节点一个来自 A 集合，一个来自 B 集合，我们就将这个图称为二分图。
+- 这一题可以用 BFS、DFS、并查集来解答。这里是 DFS 实现。任选一个节点开始，把它染成红色，然后对整个图 DFS 遍历，把与它相连的节点并且未被染色的，都染成绿色。颜色不同的节点代表不同的集合。这时候还可能遇到第 2 种情况，与它相连的节点已经有颜色了，并且这个颜色和前一个节点的颜色相同，这就说明了该无向图不是二分图。可以直接 return false。如此遍历到所有节点都染色了，如果能染色成功，说明该无向图是二分图，返回 true。
+
+## 代码
+
+```go
+package leetcode
+
+// DFS 染色，1 是红色，0 是绿色，-1 是未染色
+func isBipartite(graph [][]int) bool {
+	colors := make([]int, len(graph))
+	for i := range colors {
+		colors[i] = -1
+	}
+	for i := range graph {
+		if !dfs(i, graph, colors, -1) {
+			return false
+		}
+	}
+	return true
+}
+
+func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
+	if colors[n] == -1 {
+		if parentCol == 1 {
+			colors[n] = 0
+		} else {
+			colors[n] = 1
+		}
+	} else if colors[n] == parentCol {
+		return false
+	} else if colors[n] != parentCol {
+		return true
+	}
+	for _, c := range graph[n] {
+		if !dfs(c, graph, colors, colors[n]) {
+			return false
+		}
+	}
+	return true
+}
+```

leetcode/785.Is Graph Bipartite?

@@ -1,44 +0,0 @@
-func isBipartite(graph [][]int) bool {
-    colors := make([]int,len(graph))
-    
-    
-    for i := range colors {
-        colors[i] = -1
-    }
-
-    
-    for i := range graph {
-        if !dfs(i, graph, colors, -1) {
-            fmt.Println(colors)
-            return false
-        }
-    }
-    
-    fmt.Println(colors)
-    return true
-    
-}
-
-func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
-    if colors[n] == -1 {
-        if parentCol == 1 {
-            colors[n] = 0
-        } else {
-            colors[n] = 1 
-        }
-    } else if colors[n] == parentCol {
-        fmt.Println(n)
-        return false
-    } else if colors[n] != parentCol {
-        return true
-    }
-    
-    
-    for _, c := range graph[n] {
-        if !dfs(c, graph, colors, colors[n]) {
-            fmt.Println(c)
-            return false
-        }
-    }
-    return true
-}