資訊專欄INFORMATION COLUMN
摘要：兩個(gè)循環(huán)遍歷整個(gè)矩陣，出現(xiàn)則將其周圍相鄰的全部標(biāo)記為，用子函數(shù)遞歸標(biāo)記。注意里每次遞歸都要判斷邊界。寫一個(gè)的，寫熟練。
Number of Islands Problem
Given a boolean/char 2D matrix, find the number of islands.
0 is represented as the sea, 1 is represented as the island. If two 1 is adjacent, we consider them in the same island. We only consider up/down/left/right adjacent.
ExampleGiven graph:
[ [1, 1, 0, 0, 0], [0, 1, 0, 0, 1], [0, 0, 0, 1, 1], [0, 0, 0, 0, 0], [0, 0, 0, 0, 1] ]
return 3.
Note兩個(gè)for循環(huán)遍歷整個(gè)矩陣，出現(xiàn)"1"則count++將其周圍相鄰的"1"全部標(biāo)記為"0"，用子函數(shù)mark()遞歸標(biāo)記。
Solution DFS注意mark里每次遞歸都要判斷邊界。
public class Solution {
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
int count = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == "1") {
count++;
mark(grid, i, j);
}
}
}
return count;
}
public void mark(char[][] grid, int i, int j) {
if (grid[i][j] == "1" && i >= 0 && i < grid.length && j >= 0 && j < grid[0].length) {
grid[i][j] = "0";
if (i-1 >= 0) mark(grid, i-1, j);
if (i+1 < grid.length) mark(grid, i+1, j);
if (j-1 >= 0) mark(grid, i, j-1);
if (j+1 < grid[0].length) mark(grid, i, j+1);
}
}
}
Union Find
寫一個(gè)Union Find的class，寫熟練。
public class Solution {
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
int m = grid.length, n = grid[0].length;
UnionFind uf = new UnionFind(m, n, grid);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == "0") continue;
int x = i*n+j;
int y;
if (i < m-1 && grid[i+1][j] == "1") {
y = x+n;
uf.union(x, y);
}
if (j < n-1 && grid[i][j+1] == "1") {
y = x+1;
uf.union(x, y);
}
}
}
return uf.count;
}
class UnionFind {
public int count;
public int[] parents;
public UnionFind(int m, int n, char[][] grid) {
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == "1") count++;
}
}
parents = new int[m*n];
for (int i = 0; i < m*n; i++) parents[i] = i;
}
public int find(int i) {
if (i == parents[i]) return i;
parents[i] = find(parents[i]);
return parents[i];
}
public void union(int i, int j) {
int pi = find(i);
int pj = find(j);
if (pi == pj) return;
else parents[pi] = pj;
count--;
}
}
}
Number of Islands II
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