https://leetcode-cn.com/problems/unique-paths/

我的方法一：动态规划

步骤

1. 确认状态
   1.1 最后一步：到达第(m, n)点，或者从(m-1, n)过来，或者从(m, n-1)过来
   1.2 子问题：f(m, n)可以转化为f(m-1, n)和(m, n-1)两个问题
2. 转移方程
   f(m, n) = f(m-1, n) + f(m, n-1)
3. 初始条件和边界条件
   3.1 m>0, n>0
   3.2 f(1, 1) = 1, f(*, 0) = 0, f(0, *) = 0
4. 计算顺序
   f(1, 1), f(1,2), f(2,1) f(m, n)

复杂度

时间复杂度：O(mxn)
空间复杂度：O(mxn)

代码

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m < 1 || n<1){
            return 0;
        }

        int ** f = new int*[m+1];
        for(int i = 0; i<=m; i++){
            f[i] = new int[n+1];
            for(int j = 0; j<=n; j++){
                f[i][j] = 0;
            }
        }

        f[1][1] = 1;

        for(int i = 1; i<=m; i++){
            for(int j = 1; j<=n; j++){
                if(i>1 || j>1){
                    f[i][j] = f[i][j-1]+f[i-1][j];
                }
            }
        }

        int count = f[m][n];
        
        for(int i = 0; i<=m; i++){
            delete[] f[i];
            f[i] = 0;
        }
        delete[] f;
        f = 0;
        
        return count;
    }
};