手机微网站开发教程,黄州区精神文明建设网站,成都网站建设sntuu,网上开店创业leetcode算法题200 链接#xff1a;https://leetcode.cn/problems/number-of-islands 题目
你一个由 ‘1’#xff08;陆地#xff09;和 ‘0’#xff08;水#xff09;组成的的二维网格#xff0c;请你计算网格中岛屿的数量。
岛屿总是被水包围#xff0c;并且每座岛… 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) {StackInteger 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;}}
