leetcode: 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?

# Note: m and n will be at most 100.

DP算法。
Climbing Stairs二维版。计算解个数的题多半是用DP。

dp[i][j]表示从起点到位置(i, j)的路径总数。
DP题目定义好状态后，接下去有两个任务：找通项公式，以及确定计算的方向。

1. 由于只能向右和左走，所以对于(i, j)来说，只能从左边或上边的格子走下来：
    dp[i][j] = dp[i-1][j] + dp[i][j-1]
2. 对于网格最上边和最左边，则只能从起点出发直线走到，dp[0][j] = dp[i][0] = 1
3. 计算方向从上到下，从左到右即可。可以用滚动数组实现。
4. 为了节省空间，我们使用一维数组dp，一行一行的刷新也可以。

class Solution():
    def uniquePaths(self, m, n):
        ways = [1] * m
        for _ in range(1, n):
            for j in range(1, m):
                ways[j] += ways[j - 1]
        return ways[-1]


if __name__ == "__main__":
    assert Solution().uniquePaths(2, 1) == 1