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Add solution 0785
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35
leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite.go
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35
leetcode/0785.Is-Graph-Bipartite/785. Is Graph Bipartite.go
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@ -0,0 +1,35 @@
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package leetcode
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// DFS 染色，1 是红色，0 是绿色，-1 是未染色
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func isBipartite(graph [][]int) bool {
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colors := make([]int, len(graph))
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for i := range colors {
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colors[i] = -1
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}
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for i := range graph {
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if !dfs(i, graph, colors, -1) {
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return false
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}
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}
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return true
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}
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func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
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if colors[n] == -1 {
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if parentCol == 1 {
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colors[n] = 0
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} else {
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colors[n] = 1
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}
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} else if colors[n] == parentCol {
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return false
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} else if colors[n] != parentCol {
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return true
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}
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for _, c := range graph[n] {
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if !dfs(c, graph, colors, colors[n]) {
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return false
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}
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}
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return true
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}
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@ -0,0 +1,52 @@
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question785 struct {
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para785
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ans785
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}
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// para 是参数
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// one 代表第一个参数
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type para785 struct {
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graph [][]int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans785 struct {
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one bool
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}
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func Test_Problem785(t *testing.T) {
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qs := []question785{
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{
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para785{[][]int{{1, 3}, {0, 2}, {1, 3}, {0, 2}}},
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ans785{true},
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},
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{
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para785{[][]int{{1, 2, 3}, {0, 2}, {0, 1, 3}, {0, 2}}},
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ans785{false},
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},
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{
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para785{[][]int{{1, 2, 3}, {0, 2, 3}, {0, 1, 3}, {0, 1, 2}}},
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ans785{false},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 785------------------------\n")
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for _, q := range qs {
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_, p := q.ans785, q.para785
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fmt.Printf("【input】:%v 【output】:%v\n", p, isBipartite(p.graph))
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}
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fmt.Printf("\n\n\n")
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}
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98
leetcode/0785.Is-Graph-Bipartite/README.md
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98
leetcode/0785.Is-Graph-Bipartite/README.md
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# [785. Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite/)
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## 题目
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Given an undirected `graph`, return `true` if and only if it is bipartite.
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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.
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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: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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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: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.
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**Note:**
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- `graph` will have length in range `[1, 100]`.
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- `graph[i]` will contain integers in range `[0, graph.length - 1]`.
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- `graph[i]` will not contain `i` or duplicate values.
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- The graph is undirected: if any element `j` is in `graph[i]`, then `i` will be in `graph[j]`.
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## 题目大意
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给定一个无向图 graph，当这个图为二分图时返回 true。
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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。
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注意:
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- graph 的长度范围为 [1, 100]。
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- graph[i] 中的元素的范围为 [0, graph.length - 1]。
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- graph[i] 不会包含 i 或者有重复的值。
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- 图是无向的: 如果 j 在 graph[i] 里边, 那么 i 也会在 graph[j] 里边。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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// DFS 染色，1 是红色，0 是绿色，-1 是未染色
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func isBipartite(graph [][]int) bool {
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colors := make([]int, len(graph))
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for i := range colors {
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colors[i] = -1
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}
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for i := range graph {
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if !dfs(i, graph, colors, -1) {
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return false
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}
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}
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return true
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}
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func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
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if colors[n] == -1 {
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if parentCol == 1 {
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colors[n] = 0
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} else {
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colors[n] = 1
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}
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} else if colors[n] == parentCol {
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return false
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} else if colors[n] != parentCol {
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return true
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}
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for _, c := range graph[n] {
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if !dfs(c, graph, colors, colors[n]) {
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return false
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}
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}
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return true
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}
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```
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@ -1,44 +0,0 @@
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func isBipartite(graph [][]int) bool {
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colors := make([]int,len(graph))
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for i := range colors {
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colors[i] = -1
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}
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for i := range graph {
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if !dfs(i, graph, colors, -1) {
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fmt.Println(colors)
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return false
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}
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}
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fmt.Println(colors)
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return true
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}
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func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
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if colors[n] == -1 {
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if parentCol == 1 {
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colors[n] = 0
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} else {
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colors[n] = 1
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}
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} else if colors[n] == parentCol {
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fmt.Println(n)
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return false
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} else if colors[n] != parentCol {
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return true
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}
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for _, c := range graph[n] {
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if !dfs(c, graph, colors, colors[n]) {
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fmt.Println(c)
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return false
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}
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}
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return true
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}
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98
website/content/ChapterFour/0785.Is-Graph-Bipartite.md
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98
website/content/ChapterFour/0785.Is-Graph-Bipartite.md
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@ -0,0 +1,98 @@
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# [785. Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite/)
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## 题目
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Given an undirected `graph`, return `true` if and only if it is bipartite.
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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.
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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: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: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.
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**Note:**
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- `graph` will have length in range `[1, 100]`.
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- `graph[i]` will contain integers in range `[0, graph.length - 1]`.
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- `graph[i]` will not contain `i` or duplicate values.
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- The graph is undirected: if any element `j` is in `graph[i]`, then `i` will be in `graph[j]`.
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## 题目大意
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给定一个无向图 graph，当这个图为二分图时返回 true。
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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。
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注意:
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- graph 的长度范围为 [1, 100]。
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- graph[i] 中的元素的范围为 [0, graph.length - 1]。
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- graph[i] 不会包含 i 或者有重复的值。
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- 图是无向的: 如果 j 在 graph[i] 里边, 那么 i 也会在 graph[j] 里边。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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// DFS 染色，1 是红色，0 是绿色，-1 是未染色
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func isBipartite(graph [][]int) bool {
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colors := make([]int, len(graph))
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for i := range colors {
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colors[i] = -1
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}
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for i := range graph {
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if !dfs(i, graph, colors, -1) {
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return false
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}
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}
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return true
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}
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func dfs(n int, graph [][]int, colors []int, parentCol int) bool {
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if colors[n] == -1 {
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if parentCol == 1 {
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colors[n] = 0
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} else {
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colors[n] = 1
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}
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} else if colors[n] == parentCol {
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return false
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} else if colors[n] != parentCol {
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return true
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}
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for _, c := range graph[n] {
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if !dfs(c, graph, colors, colors[n]) {
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return false
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}
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}
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return true
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}
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```
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@ -379,6 +379,7 @@ headless: true
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- [0778.Swim-in-Rising-Water]({{< relref "/ChapterFour/0778.Swim-in-Rising-Water.md" >}})
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- [0781.Rabbits-in-Forest]({{< relref "/ChapterFour/0781.Rabbits-in-Forest.md" >}})
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- [0784.Letter-Case-Permutation]({{< relref "/ChapterFour/0784.Letter-Case-Permutation.md" >}})
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- [0785.Is-Graph-Bipartite]({{< relref "/ChapterFour/0785.Is-Graph-Bipartite.md" >}})
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- [0786.K-th-Smallest-Prime-Fraction]({{< relref "/ChapterFour/0786.K-th-Smallest-Prime-Fraction.md" >}})
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- [0793.Preimage-Size-of-Factorial-Zeroes-Function]({{< relref "/ChapterFour/0793.Preimage-Size-of-Factorial-Zeroes-Function.md" >}})
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- [0802.Find-Eventual-Safe-States]({{< relref "/ChapterFour/0802.Find-Eventual-Safe-States.md" >}})
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