LeetCode OJ Number of Islands
来源：程序员人生 发布时间：2015-04-21 08:51:11 阅读次数：3073次
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, ⑴, 0, 0, 1, 0, ⑴ };
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;
}
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