[LeetCode每日一题]62. Unique Paths

题目如下：

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?

62. Unique Paths
给定一个mxn的矩阵，一机器人从左上角grid[0][0]出发，该机器人只能向下走或者向右走，要到达终点右下角grid[m-1][n-1],问最多有多少种不同的走法。
这道题目是典型的动态规划问题，设当前机器人处在grid[x][y],则之前机器人可能是向右走或向下走到达grid[x][y],设机器人处在grid[i][j]时的走法个数为dp[i][j],则dp[i][j] = dp[i-1][j] + dp[i][j-1]。
此外还应考虑边界条件dp[i][0]==0和dp[0][j]==0。
实现代码如下：

class Solution {
    public int uniquePaths(int m, int n) {
        if (m <= 1 || n <= 1) {
            return 1;
        }
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n; j++) {
            dp[0][j] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}