LeetCode 200：岛屿数量，把网格陆地看成连通分量
题目要求 给你一个 m x n 的二维字符网格 grid： "1" 表示陆地 "0" 表示水 题目要求返回岛屿数量。 一个岛屿由水平或垂直相邻的陆地组成。对角线相邻不算连通。可以认为网格四周都被水包围。 输入输出 输入：grid: List[List[str]] 输出：岛屿数量 int 只看上下左右四个方向。 "0" 水格子不能算作岛屿的一部分。 示例 输入： [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ] 输出：1 这些陆地通过上下左右连成一整块，所以答案是 1。 输入： [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ] 输出：3 这里有三块互不连通的陆地，所以答案是 3。 约束 m == grid.length n == grid[i].length 1 <= m, n <= 300 grid[i][j] 只会是 "0" 或 "1" 这一题可以用 DFS 或 BFS 做。这里的目标是练并查集：把每块陆地当成一个节点，把相邻陆地合并，最后留下的陆地连通分量数量就是岛屿数量。 ...