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Commit 7cf45ae

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feat: add js solution to lc problem: No.1905 (doocs#3453)
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‎solution/1900-1999/1905.Count Sub Islands/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -41,7 +41,7 @@ grid2 中标红的 1 区域是子岛屿,总共有 3 个子岛屿。
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<p><strong>示例 2:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1905.Count%20Sub%20Islands/images/testcasex2.png" style="width: 491px; height: 201px;">
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<pre><b>输入:</b>grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
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<b>输出:</b>2
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<b>输出:</b>2
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<strong>解释:</strong>如上图所示,左边为 grid1 ,右边为 grid2 。
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grid2 中标红的 1 区域是子岛屿,总共有 2 个子岛屿。
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</pre>
@@ -220,6 +220,35 @@ function countSubIslands(grid1: number[][], grid2: number[][]): number {
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}
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```
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#### JavaScript
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```js
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function countSubIslands(grid1, grid2) {
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const [m, n] = [grid1.length, grid1[0].length];
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let ans = 0;
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const dirs = [-1, 0, 1, 0, -1];
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const dfs = (i, j) => {
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let ok = grid1[i][j];
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grid2[i][j] = 0;
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for (let k = 0; k < 4; ++k) {
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const [x, y] = [i + dirs[k], j + dirs[k + 1]];
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if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y]) {
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ok &= dfs(x, y);
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}
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}
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return ok;
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};
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for (let i = 0; i < m; ++i) {
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for (let j = 0; j < n; j++) {
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if (grid2[i][j]) {
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ans += dfs(i, j);
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}
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1900-1999/1905.Count Sub Islands/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -42,7 +42,7 @@ The 1s colored red in grid2 are those considered to be part of a sub-island. The
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1905.Count%20Sub%20Islands/images/testcasex2.png" style="width: 491px; height: 201px;" />
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<pre>
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<strong>Input:</strong> grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
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<strong>Output:</strong> 2
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<strong>Output:</strong> 2
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<strong>Explanation: </strong>In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
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The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.
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</pre>
@@ -220,6 +220,35 @@ function countSubIslands(grid1: number[][], grid2: number[][]): number {
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}
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```
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#### JavaScript
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```js
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function countSubIslands(grid1, grid2) {
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const [m, n] = [grid1.length, grid1[0].length];
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let ans = 0;
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const dirs = [-1, 0, 1, 0, -1];
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const dfs = (i, j) => {
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let ok = grid1[i][j];
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grid2[i][j] = 0;
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for (let k = 0; k < 4; ++k) {
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const [x, y] = [i + dirs[k], j + dirs[k + 1]];
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if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y]) {
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ok &= dfs(x, y);
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}
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}
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return ok;
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};
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for (let i = 0; i < m; ++i) {
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for (let j = 0; j < n; j++) {
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if (grid2[i][j]) {
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ans += dfs(i, j);
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}
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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function countSubIslands(grid1, grid2) {
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const [m, n] = [grid1.length, grid1[0].length];
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let ans = 0;
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const dirs = [-1, 0, 1, 0, -1];
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const dfs = (i, j) => {
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let ok = grid1[i][j];
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grid2[i][j] = 0;
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for (let k = 0; k < 4; ++k) {
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const [x, y] = [i + dirs[k], j + dirs[k + 1]];
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if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y]) {
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ok &= dfs(x, y);
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}
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}
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return ok;
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};
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for (let i = 0; i < m; ++i) {
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for (let j = 0; j < n; j++) {
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if (grid2[i][j]) {
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ans += dfs(i, j);
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}
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}
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}
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return ans;
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}

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