62.不同路径

力扣题目链接

代码

示例代码

class Solution {
public:int uniquePaths(int m, int n) {vector<vector<int>> result(m, vector<int>(n, 0));for (int i = 0; i < n; i++) {result[0][i] = 1;}for (int i = 0; i < m; i++) {result[i][0] = 1;}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {result[i][j] = result[i - 1][j] + result[i][j - 1];}}return result[m-1][n-1];}
};

思路

从前做了，还算有印象，当前元素从上边和左边的值计算得出

63. 不同路径 II

力扣题目链接

代码

示例代码

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];}
};

思路

加入障碍后，代码需要相应地改变