leetcode算法题基础（三十七） 并查集（一）200 岛屿数量

这道题的输入是一个二维数组(n ∗ m n*mn∗m)，然后要我们求联通区域的块数
我的思路是创建一个一个长度为n ∗ m n*mn∗m的一维数组作为初始并查集，然后使用遍历输入的二维数组，每当发现某个位置的右边或者下面的值是1，并且本身的值也是1的时候，合并这两个区域。（注意二维数组的索引不要搞混，我写的时候搞混导致调试了很久）
代码如下

class Solution(object):
    #创建初始并查集
    def __init__(self):
        self.ans = [i for i in range(100000)]
    #查
    def find(self, x):
        while self.ans[x] != x:
            x = self.ans[x]
        return x
    #并
    def merge(self, i, j):
        fi = self.find(i)
        fj = self.find(j)
        if self.ans[fi] != self.ans[fj]:
            self.ans[fi] = fj

    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                t1 = i*len(grid[0])+j
                t2 = t1+1
                t3 = t1+len(grid[0])
                if j+1<len(grid[0]) and grid[i][j] == "1" and grid[i][j+1] == "1":
                    self.merge(t1, t2)

                if i+1<len(grid) and grid[i][j] == "1" and grid[i+1][j] == "1":
                    self.merge(t1, t3)

                # print(self.ans[:(len(grid)*len(grid[0]))])
                
        answer = 0
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if self.ans[i*len(grid[0])+j] == i*len(grid[0])+j and grid[i][j] == "1":
                    answer += 1
        return answer
        
# s = Solution()
# ans = s.numIslands(
# [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]])