leetcode算法题200
链接：https://leetcode.cn/problems/number-of-islands

题目

你一个由 ‘1’（陆地）和 ‘0’（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [
[“1”,“1”,“1”,“1”,“0”],
[“1”,“1”,“0”,“1”,“0”],
[“1”,“1”,“0”,“0”,“0”],
[“0”,“0”,“0”,“0”,“0”]
]
输出：1
示例 2：

输入：grid = [
[“1”,“1”,“0”,“0”,“0”],
[“1”,“1”,“0”,“0”,“0”],
[“0”,“0”,“1”,“0”,“0”],
[“0”,“0”,“0”,“1”,“1”]
]
输出：3

解法

1、感染

	public static int numIslands(char[][] grid) {int num = 0;int X = grid.length;for (int i = 0; i < X; i++) {int Y = grid[i].length;for (int j = 0; j < Y; j++) {if (grid[i][j] == '1') {num++;// 填充设置岛屿infect(grid, i, j);}}}return num;}public static void infect(char[][] grid, int x, int y) {if (x < 0 || x == grid.length || y < 0 || y == grid[0].length || grid[x][y] != '1') {return;}grid[x][y] = '0';infect(grid, x - 1, y);infect(grid, x + 1, y);infect(grid, x, y - 1);infect(grid, x, y + 1);}

2、并查集

和之前的省份数量有点类似，这里取巧了一下，使用一维数组来存储所有情况，通过x * col + y算出下标

	public static int numIslands2(char[][] grid) {if (null == grid || grid.length == 0) {return 0;}int row = grid.length;int col = grid[0].length;UnionFind unionFind = new UnionFind(grid);// 0列特殊处理，后面不用判断边界for (int i = 1; i < row; i++) {if (grid[i - 1][0] == '1' && grid[i][0] == '1') {unionFind.union(i - 1, 0, i, 0);}}// 0行特殊处理，后面不用判断边界for (int i = 1; i < col; i++) {if (grid[0][i - 1] == '1' && grid[0][i] == '1') {unionFind.union(0, i - 1, 0, i);}}for (int i = 1; i < grid.length; i++) {for (int j = 1; j < grid[i].length; j++) {if (grid[i][j] == '1') {if (grid[i - 1][j] == '1') {unionFind.union(i, j, i - 1, j);}if (grid[i][j - 1] == '1') {unionFind.union(i, j, i, j - 1);}}}}return unionFind.getSize();}public static class UnionFind {private int[] parents;private int[] childSizes;/*** 每行有多少个*/private int col;/*** 一共有多少个集合（岛）*/private int size;public UnionFind(char[][] data) {int row = data.length;col = data[0].length;int len = row * col;parents = new int[len];childSizes = new int[len];for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {if (data[i][j] == '1') {int index = getIndex(i, j);parents[index] = index;childSizes[index] = 1;size++;}}}}//	坐标换成点 （x,y)->pointpublic int getIndex(int x, int y) {return x * col + y;}public int findParent(int index) {Stack<Integer> stack = new Stack<>();while (index != parents[index]) {stack.push(index);index = parents[index];}while (!stack.isEmpty()) {parents[stack.pop()] = index;}return index;}public void union(int x1, int y1, int x2, int y2) {int index1 = getIndex(x1, y1);int index2 = getIndex(x2, y2);int parent1 = findParent(index1);int parent2 = findParent(index2);if (parent1 != parent2) {if (childSizes[parent1] >= childSizes[parent2]) {childSizes[parent1] += childSizes[parent2];parents[parent2] = parent1;} else {childSizes[parent2] += childSizes[parent1];parents[parent1] = parent2;}size--;}}public int getSize() {return size;}}