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`‎problems/785.is-graph-bipartite.md`

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`@@ -49,7 +49,11 @@ graph[i] 不会包含 i 或者有重复的值。`
`49`	`49`
`50`	`50`	`- 暂无`
`51`	`51`
`52`		`-## 思路`
	`52`	`+## 着色法 + DFS`
	`53`	`+`
	`54`	`+求二分图有两种思路,一个是着色法,另外一个是并查集。`
	`55`	`+`
	`56`	`+### 思路`
`53`	`57`
`54`	`58`	`和 886 思路一样。我甚至直接拿过来 dfs 函数一行代码没改就 AC 了。`
`55`	`59`
`@@ -68,12 +72,12 @@ graph[i] 不会包含 i 或者有重复的值。`
`68`	`72`
`69`	`73`	`强烈建议两道题一起练习一下。`
`70`	`74`
`71`		`-## 关键点`
	`75`	`+### 关键点`
`72`	`76`
`73`	`77`	`- 图的建立和遍历`
`74`	`78`	`- colors 数组`
`75`	`79`
`76`		`-## 代码`
	`80`	`+### 代码`
`77`	`81`
`78`	`82`	```py
`79`	`83`	`class Solution:`
`@@ -102,8 +106,10 @@ class Solution:`
`102`	`106`
`103`	`107`	`复杂度分析`
`104`	`108`
`105`		`-- 时间复杂度:$O(N^2)$`
`106`		`-- 空间复杂度:$O(N)$`
	`109`	`+令 v 和 e 为图中的顶点数和边数。`
	`110`	`+`
	`111`	`+- 时间复杂度:$O(v+e)$`
	`112`	`+- 空间复杂度:$O(v+e)$`
`107`	`113`
`108`	`114`
`109`	`115`	`如上代码并不优雅,之所以这么写只是为了体现和 886 题一致性。一个更加优雅的方式是不建立 grid,而是利用题目给的 graph(邻接矩阵)。`
`@@ -123,6 +129,47 @@ class Solution:`
`123`	`129`	`if colors[i] == 0 and not dfs(i,1): return False`
`124`	`130`	`return True`
`125`	`131`	```
	`132`	`+## 并查集`
	`133`	`+`
	`134`	`+### 思路`
	`135`	`+`
	`136`	`+遍历图,对于每一个顶点 i,将其所有邻居进行合并,合并到同一个联通域中。这样当发现某个顶点 i 和其邻居已经在同一个联通分量的时候可以直接返回 false,否则返回 true。`
	`137`	`+`
	`138`	`+### 代码`
	`139`	`+`
	`140`	+```py
	`141`	`+class UF:`
	`142`	`+ def __init__(self, n):`
	`143`	`+ self.parent = {}`
	`144`	`+ for i in range(n):`
	`145`	`+ self.parent[i] = i`
	`146`	`+ def union(self, i,j):`
	`147`	`+ self.parent[self.find(i)] = self.find(j)`
	`148`	`+ def find(self, i):`
	`149`	`+ if i == self.parent[i]: return i`
	`150`	`+ self.parent[i] = self.find(self.parent[i])`
	`151`	`+ return self.parent[i]`
	`152`	`+ def is_connected(self, i,j):`
	`153`	`+ return self.find(i) == self.find(j)`
	`154`	`+`
	`155`	`+class Solution:`
	`156`	`+ def isBipartite(self, graph: List[List[int]]) -> bool:`
	`157`	`+ n = len(graph)`
	`158`	`+ uf = UF(n)`
	`159`	`+ for i in range(n):`
	`160`	`+ for neibor in graph[i]:`
	`161`	`+ if uf.is_connected(i, neibor): return False`
	`162`	`+ uf.union(graph[i][0], neibor)`
	`163`	`+ return True`
	`164`	+```
	`165`	`+`
	`166`	`+`
	`167`	`+复杂度分析`
	`168`	`+`
	`169`	`+令 v 和 e 为图中的顶点数和边数。`
	`170`	`+`
	`171`	`+- 时间复杂度:$O(v+e)$`
	`172`	`+- 空间复杂度:$O(v+e)$`
`126`	`173`
`127`	`174`	`## 相关问题`
`128`	`175`

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