【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).
How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2Output: 3Explanation:From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:1. Right -> Right -> Down2. Right -> Down -> Right3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3Output: 28
【题意】
从格子的左上角走到格子的右下角一共有多少种走法。
【解答】
使用二维矩阵来存储路径数目。在每一个点，要么从上方走来，要么从左边走来，所以matrix[i][j] = matrix[i-1][j] + matrix[i][j-1]。最上面一行以及最左边一列的走法数目都为1，如此更新matrix即可。
递归的话一定会超时。
时间O(m*n) 空间O(m*n)。
class Solution {public: int uniquePaths(int m, int n) { long long matrix[n][m] = { 0}; for(int i=0; i