【题目】

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).

Note: m and n will be at most 100.

题意：数组 A[m][n] ，从 A[0][0] 到 A[m-1][n-1] 有多少条路径。

对动态规划不熟的同学，可以借这个例子由浅入深理解一下。

二维数组实现：

public class Solution {
    public int uniquePaths(int m, int n) {
        // DP with 2 dimensions array
        int[][] a = new int[m][n];
        for (int i = 0; i < m; i++) {
            a[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            a[0][i] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                a[i][j] = a[i-1][j] + a[i][j-1];
            }
        }
        return a[m-1][n-1];
    }
}

一维数组实现：

public class Solution {
    public int uniquePaths(int m, int n) {
        // DP with 1 dimension array
        int[] a = new int[n];
        for (int j = 0; j < n; j++) {
            a[j] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                a[j] += a[j-1];
            }
        }
        return a[n-1];
    }
}

一维数组升级版：

public class Solution {
    public int uniquePaths(int m, int n) {
        // Advanced: DP with 1 dimension array
    	int row = Math.min(m, n);
        int col = Math.max(m, n);
        int[] a = new int[col];
        for (int j = 0; j < col; j++) {
            a[j] = 1;
        }
        for (int i = 1; i < row; i++) {
            a[i] *= 2;
            for (int j = i+1; j < col; j++) {
                a[j] += a[j-1];
            }
        }
        return a[col-1];
    }
}

【LeetCode】Unique Paths 解题报告