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
题目大意：
从m乘n矩阵左上角走到右下角，每一步只能向右或向下，求共有多少条不同路径。
递归解决：
1 class Solution { 2 public: 3 vector<vector<int>> v; 4 5 int solve(int m, int n) {//从(m, n)到(1, 1)有多少条路径 6 if (m == 1 || n == 1) 7 return 1; 8 if (v[m][n] >= 0) 9 return v[m][n]; 10 v[m][n] = solve(m - 1, n) + solve(m, n - 1); 11 return v[m][n]; 12 } 13 14 int uniquePaths(int m, int n) { 15 if (m == 0 || n == 0) 16 return 0; 17 if (m == 1 || n == 1) 18 return 1; 19 v.resize(m + 1, vector<int>(n + 1, -1)); 20 return solve(m - 1, n) + solve(m, n - 1); 21 } 22 };
迭代：
1 class Solution { 2 public: 3 4 int uniquePaths(int m, int n) { 5 if (m == 0 || n == 0) 6 return 0; 7 vector<vector<int>> v(m, vector<int>(n)); 8 int i, j; 9 for (i = 0; i < m; i++) { 10 for (j = 0; j < n; j++) { 11 if (i == 0 || j == 0) 12 v[i][j] = 1; 13 else 14 v[i][j] = v[i - 1][j] + v[i][j - 1]; 15 } 16 } 17 return v[m - 1][n - 1]; 18 } 19 };
原文：https://www.cnblogs.com/lxc1910/p/10482789.html