代码随想录算法训练营第37期 第三十九天 | LeetCode62.不同路径、63. 不同路径 II
一、62.不同路径
解题代码C++：
class Solution {
public:int uniquePaths(int m, int n) {vector<vector<int>> dp(m, vector<int>(n, 0));for (int i = 0; i < m; i++) dp[i][0] = 1;for (int j = 0; j < n; j++) dp[0][j] = 1;for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}return dp[m - 1][n - 1];}
};
题目链接/文章讲解/视频讲解：
https://programmercarl.com/0062.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84.html
二、63. 不同路径 II
解题代码C++：
class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m = obstacleGrid.size();int n = obstacleGrid[0].size();if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍，直接返回0return 0;vector<vector<int>> dp(m, vector<int>(n, 0));for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {if (obstacleGrid[i][j] == 1) continue;dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}return dp[m - 1][n - 1];}
};
题目链接/文章讲解/视频讲解：
https://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.html
