62. Unique Paths

####问题描述
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?

Above is a 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:

1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

题目链接：

####思路分析
m×n的地图，机器人智能向下或者向右走，计算不同的从左上到右下的路径。

简单的动态规划问题，建一个二维数组，到达每一个位置的路径数等于它上面的位置加左边的位置。如果是在边界上，则为零。res[0][0] = 1，顺序遍历计算返回finish位置的值即可。
####代码
java

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        dp[1][1] = 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][n];     
    }
}

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> res(m, vector<int>(n, 0));
        res[0][0] = 1;
        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                int left = j - 1 >= 0 ? res[i][j - 1] : 0;
                int up = i - 1 >= 0 ? res[i - 1][j] : 0;
                res[i][j] = res[i][j] + left + up;
            }
        }
        return res[m - 1][n - 1];
    }
};

时间复杂度：$O(m\times n)$
空间复杂度：$O(m\times n)$

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####反思
很简单的DP，空间换时间的思想体现的淋漓尽致。