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-rw-r--r--num_closed_islands.cpp61
1 files changed, 0 insertions, 61 deletions
diff --git a/num_closed_islands.cpp b/num_closed_islands.cpp
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--- a/num_closed_islands.cpp
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-// 1254. Number of Closed Islands
-#include "leetcode.h"
-
-/*
- * given 2d 'grid' consists of 0's (land) and 1's (water). an island is a
- * maximal four directionally connected group of 0's and a closed island is an
- * island totally (all left, top, right, bottom) surrouned by 1's. return the
- * number of closed islands.
- */
-
-class Solution {
- int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
- bool is_valid(vvd & grid, int i, int j, int r, int c) {
- if (i >= 0 && i < r && j >= 0 && j < c && grid[i][j] == 0)
- return true;
- else
- return false;
- }
- void dfs(vvd & grid, int i, int j, int r, int c) {
- grid[i][j] = 1;
- for (int x = 0; x < 4; ++x) {
- int new_x = i + dx[x];
- int new_y = i + dy[x];
- if (is_valid(grid, new_x, new_y, r, c))
- dfs(grid, new_x, new_y, r, c);
- }
- }
-
-public:
- int closedIsland(vvd & grid) {
- int r = grid.size(), c = grid[0].size(), ans = 0;
- for (int i = 0; i < r; ++i)
- for (int j = 0; j < c; ++j)
- if (i * j == 0 || i == r - 1 || j == c - 1)
- if (grid[i][j] == 0)
- dfs(grid, i, j, r, c);
- for (int i = 0; i < r; ++i)
- for (int j = 0; j < c; ++j)
- if (grid[i][j] == 0) {
- ans++;
- dfs(grid, i, j, r, c);
- }
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vvd g1 = {{1, 1, 1, 1, 1, 1, 1, 0},
- {1, 0, 0, 0, 0, 1, 1, 0},
- {1, 0, 0, 0, 0, 1, 0, 1},
- {1, 1, 1, 1, 1, 1, 1, 0}};
- vvd g2 = {{0, 0, 1, 0, 0}, {0, 1, 0, 1, 0}};
- vvd
- g3 = {{1, 1, 1, 1, 1, 1, 1}, {1, 0, 0, 0, 0, 0, 1}, {1, 0, 1, 1, 1, 0, 1},
- {1, 0, 1, 0, 1, 0, 1}, {1, 0, 1, 1, 1, 0, 1}, {1, 0, 0, 0, 0, 0, 1},
- {1, 1, 1, 1, 1, 1, 1}};
- printf("%d\n", obj.closedIsland(g1)); // expect: 2
- printf("%d\n", obj.closedIsland(g2)); // expect: 1
- printf("%d\n", obj.closedIsland(g3)); // expect: 2
-}