62. Unique Paths

【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?

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<n; i++){
            matrix[i][0] = 1;
        }
        for(int j=0; j<m; j++){
            matrix[0][j] = 1;
        }
        for(int i=1; i<n; i++){
            for(int j=1; j<m; j++){
                matrix[i][j] = matrix[i-1][j] + matrix[i][j-1];
            }
        }
        return matrix[n-1][m-1];
    }
};