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| 1 | +class Solution { |
| 2 | + int count; |
| 3 | + public int uniquePaths(int m, int n) { |
| 4 | + /** |
| 5 | + 1. dfs + backtrack => Time Limit Exceeded |
| 6 | + */ |
| 7 | + //boolean[][] visited = new boolean[m][n]; |
| 8 | + //dfs(visited, 0, 0, m, n); |
| 9 | + |
| 10 | + /** |
| 11 | + 2. DP |
| 12 | + - ์ ์ฅ๋๋ ๊ฐ์ ํด๋น ์์น๊น์ง ๋๋ฌํ ์ ์๋ '๊ณ ์ ํ ๊ฒฝ๋ก์ ์' |
| 13 | + */ |
| 14 | + int[][] dp = new int[m][n]; |
| 15 | + |
| 16 | + for (int i = 0; i < m; i++) { |
| 17 | + dp[i][0] = 1; |
| 18 | + } |
| 19 | + for (int j = 0; j < n; j++) { |
| 20 | + dp[0][j] = 1; |
| 21 | + } |
| 22 | + |
| 23 | + for (int i = 1; i < m; i++) { |
| 24 | + for (int j = 1; j < n; j++) { |
| 25 | + // ์ด์ ์ ์๋ ๊ฐ๋ค์ ๋ํจ(์, ์ผ์ชฝ) |
| 26 | + dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; |
| 27 | + } |
| 28 | + } |
| 29 | + |
| 30 | + return dp[m-1][n-1]; |
| 31 | + } |
| 32 | + |
| 33 | + /** |
| 34 | + 1. DFS + backtrack |
| 35 | + */ |
| 36 | + /** |
| 37 | + public void dfs(boolean[][] visited, int x, int y, int m, int n) { |
| 38 | +
|
| 39 | + if (x == m - 1 && y == n - 1) { |
| 40 | + count++; |
| 41 | + return; |
| 42 | + } |
| 43 | +
|
| 44 | +
|
| 45 | + visited[x][y] = true; |
| 46 | +
|
| 47 | + // ์ค๋ฅธ์ชฝ์ผ๋ก ์ด๋ + backtrack |
| 48 | + if (y + 1 < n && !visited[x][y + 1]) { |
| 49 | + dfs(visited, x, y + 1, m , n); |
| 50 | + } |
| 51 | +
|
| 52 | + // ์๋์ชฝ์ผ๋ก ์ด๋ + backtrack |
| 53 | + if (x + 1 < m && !visited[x + 1][y]) { |
| 54 | + dfs(visited, x + 1, y, m, n); |
| 55 | + } |
| 56 | +
|
| 57 | + visited[x][y] = false; |
| 58 | + } |
| 59 | + */ |
| 60 | +} |
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