A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).優化
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).spa
How many possible unique paths are there?code
Above is a 7 x 3 grid. How many possible unique paths are there?blog
Note: m and n will be at most 100.it
Example 1:io
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:class
Input: m = 7, n = 3 Output: 28
解答:grid
典型的動態規劃問題,用一個矩陣來記錄方法數量,每一個數字表示從起始位置走到該位置有多少種方法。第一行和第一列均初始化爲1,每一格數字更新結果爲該格子上方和左方格子求和,表示我nums[i][j] = nums[i-1][j] + nums[i][j-1]。應該還有一些優化的方法,暫時先不討論。方法
代碼:im
1 class Solution { 2 public: 3 int uniquePaths(int m, int n) { 4 vector<vector<int>> num(m, vector<int>(n, 1)); 5 for (int i = 1; i < m; i++) 6 for (int j = 1; j < n; j++) 7 num[i][j] = num[i - 1][j] + num[i][j-1]; 8 return num[m - 1][n - 1]; 9 } 10 };
時間複雜度:O(m*n)
空間複雜度:O(m*n)