62. 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?
Above is a 7 x 3 grid. How many possible unique paths are there?

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

Constraints:

• 1 <= m, n <= 100
• It’s guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.

解题思路：

1. 确定状态
   开数组的意义 dy[m-1][n-1]
   确定状态需要两个意识：
   • 最后一步
     dy[m-1][n-1] = dy[m-2][n-1] + dy[m-1][n-2]
   • 子问题
     从后往前推子问题，问题规模渐渐变小
2. 转移方程
   dp[i][j] = dp[i-1][j] + dp[i][j-1]
3. 初始条件和边界情况
   d[0][j] = 1
   d[i][0] = 1
4. 计算顺序
   按行列的方向

class Solution(object):
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0 for i in range(n)] for i in range(m)]
        for i in range(m):
            for j in range(n):
                if i == 0:
                    dp[i][j] = 1
                elif j == 0:
                    dp[i][j] = 1
                else:
                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
        return dp[m-1][n-1]