There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

分析:

DFS深度搜索,遇见一个没有见过的,mark as visited。

然后立刻DFS把他附近的也mark掉,ans+=1

Solution

def findCircleNum(self, isConnected):
“””
:type isConnected: List[List[int]]
:rtype: int
“””
“””
DFS the matrix, whenever we run into to a 1, run DFS and clear that island with ans + 1
“””
ans = 0
rowLen = len(isConnected)

visited = [False]*rowLen
def dfs(i):
visited[i] = True
for j in range(rowLen):
if isConnected[i][j] == 1 and not visited[j]:
dfs(j)

for i in range(rowLen):
if not visited[i]:
dfs(i)
ans += 1

return ans

iteratively可以用stack来做