Given a n,m which means the row and column of the 2D matrix and an array of pair A( size k). Originally, the 2D matrix is all 0 which means there is only sea in the matrix. The list pair has k operator and each operator has two integer A[i].x, A[i].y means that you can change the grid matrix[A[i].x][A[i].y] from sea to island. Return how many island are there in the matrix after each operator.
先给出二维矩阵的行列值，之后再给出一系列操作数目，每个操作是在一个位置插入岛屿。和Number of Islands这种先给出全部岛屿布图不一样，这种情景下更适合使用并查集，因为预先给出岛屿分布，还是无法确定操作的数目，需要扫描二维矩阵，判断为1的地方进行操作。反而使用DFS更方便一些。这题每次加入一个一个岛屿相当于新加入一个集合。之后判断这个新加岛屿上下左右是否有岛屿，如果存在的话，进行集合的合并，把当前岛屿加入，并减少岛屿的个数。
# Definition for a point.
# class Point:
# def __init__(self, a=0, b=0):
# self.x = a
# self.y = b
class Solution:
# @param {int} n an integer
# @param {int} m an integer
# @param {Pint[]} operators an array of point
# @return {int[]} an integer array
def numIslands2(self, n, m, operators):
if not m or not n or not operators:
return []
UF = UnionFind()
res = []
for p in map(lambda a: (a.x, a.y), operators):
UF.add(p)
for dx in (1, 0), (0, 1), (-1, 0), (0, -1):
q = (dx[0] + p[0], dx[1]+p[1])
if q in UF.id:
UF.union(p,q)
res += [UF.count]
return res
class UnionFind(object):
def __init__(self):
self.id = {} #object
self.sz = {} #object‘weight, size here.
self.count = 0
def add(self, p):
self.id[p] = p
self.sz[p] = 1
self.count += 1
def find(self, i):
while i!= self.id[i]:
self.id[i] = self.id[self.id[i]] ＃路径压缩
i = self.id[i]
return i
def union(self, p, q):
i = self.find(p)
j = self.find(q)
if i == j:
return
if self.sz[i] > self.sz[j]: #union is only operated on heads按秩合并
i, j = j, i
self.id[i] = j
self.sz[i] += self.sz[j]
self.count -= 1
以上操作的最多合并次数为4k次，object k个。所以时间复杂度为O((4k+k)log*k).log*为迭代次数，不超过5，所以时间复杂度为O(k)大小。空间复杂度为两个dict（权重和father），O(k)大小。由于key值是tuple，空间消耗非常大，所以这种解法在lintcode上MLE. 可以每次将二维坐标转化为一维坐标来做键值。