LeetCode OJ Number of Islands

标签：leetcode   dfs   

Given a 2d grid map of ‘1‘s (land) and‘0‘s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

int R, C;
int dir[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };
bool ** vis;
char ** G;

void dfs(int pi, int pj) {
	vis[pi][pj] = true;
	int npi, npj;
	for (int i = 0; i < 4; i++) {
		npi = pi + dir[i][0];
		npj = pj + dir[i][1];
		if (0 <= npi && npi < R && 0 <= npj && npj < C && !vis[npi][npj] && G[npi][npj] == '1') dfs(npi, npj);
	}
}

int numIslands(char **grid, int numRows, int numColumns) {
	G = grid;
	vis = (bool **)malloc(sizeof(bool*) * numRows);
	for (int i = 0; i < numRows; i++) vis[i] = (bool *)malloc(sizeof(bool) * numColumns);
	for (int i = 0; i < numRows; i++)
		for (int j = 0; j < numColumns; j++) vis[i][j] = false;
	int ans = 0;
	R = numRows;
	C = numColumns;
	for (int i = 0; i < numRows; i++)
		for (int j = 0; j < numColumns; j++)
			if (!vis[i][j] && G[i][j] == '1') {
				ans++;
				dfs(i, j);
			}
	for (int i = 0; i < numRows; i++) free(vis[i]);
	free(vis);
	return ans;
}

LeetCode OJ Number of Islands

标签：leetcode   dfs   

原文地址：http://blog.csdn.net/u012925008/article/details/44954703