Unique Paths

https://leetcode.com/problems/unique-paths/

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). How many possible unique paths are there?

Thoughts

给定M*N的矩阵,

设Fi,j为从(0, 0)处往下或往右走到(i, j)处的unique path的数目. 求F(M-1, N-1).

观察到要到达(i, j)只有从(i - 1, j)和(i, j - 1)两条路，也就是说f[i][j] = f[i - i][j] + f[i][j - 1]

Code

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] f = new int[m][n];

        for (int i = 0; i < n; i++) {
           f[0][i] = 1;
        }
        for (int i = 0; i < m; i++) {
           f[i][0] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                f[i][j] = f[i - 1][j] + f[i][j - 1];
            }
        }

        return f[m - 1][n - 1];
    }
}

Analysis

TC: O(mn)

Ver.2

观察到f实际上只和当前行和上一行有关, 因此f可以只用一维数组来节省空间. f[j]在更新前表示当前行的前一行的第j列的结果, f[j] += f[j - 1]意味着上一行第j列结果和当前行第j-1列结果相加.

class Solution {
    public int uniquePaths(int m, int n) {
        int[] f = new int[n];
        f[0] = 1;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                f[j] += j > 0 ? f[j - 1] : 0;
            }
        }

        return f[n - 1];
    }
}