BFS in Graph and BFS in Matrix
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图Graph
N个点,M条边
M最大是O(N^2)的级别图上BFS时间复杂度= O(N + M)
说是O(M)问题也不大,因为M一般都比N大所以最坏情况可能是O(N^2)
矩阵Matrix
R行C列
R*C个点,R*C*2条边(每个点上下左右4条边,每条边被2个点共享)。矩阵中BFS时间复杂度= O(R * C)
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