LeetCode Online Judge 题目C# 练习 - Unique Paths
2012-10-23 00:21
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M*N grids from top-left to bottom-right find all the paths.
代码分析:
递归做法。
代码分析:
DP做法,
例如 M = 3, N = 4
其实应该还有一种方法,C(m - 1 + n - 1, m - 1),就是在5步里面,找出2步是往下走的。
C(x, y) = x! / (y! * (x- y)!), 但是因为x!太大,很容易overflow,所以还是别用了。
public static int UniquePaths(int m, int n) { int ret = 0; FindAllPaths(m, n, ref ret); return ret; } public static void FindAllPaths(int m, int n, ref int numofpaths) { if (m == 1 && n == 1) numofpaths++; else { if (m > 1) FindAllPaths(m - 1, n, ref numofpaths); if (n > 1) FindAllPaths(m, n - 1, ref numofpaths); } }
代码分析:
递归做法。
public static int UniquePathsDP(int m, int n) { int[,] grid = new int[m, n]; grid[0,0] = 1; for (int i = 1; i < m; i++) grid[i, 0] = 1; for (int i = 1; i < n; i++) grid[0, i] = 1; for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { grid[i, j] = grid[i - 1, j] + grid[i, j - 1]; } } return grid[m - 1, n - 1]; }
代码分析:
DP做法,
例如 M = 3, N = 4
1 | 1 | 1 | 1 |
1 | 2 | 3 | 4 |
1 | 3 | 6 | 10 |
C(x, y) = x! / (y! * (x- y)!), 但是因为x!太大,很容易overflow,所以还是别用了。
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