62.不同路径

力扣题目链接

class Solution {
public:
    // 深度优先搜索，问题转化为求二叉树叶子节点的个数
    int dfs(int i, int j, int m, int n) {
        if (i > m || j > n) {
            return 0;
        }
        if (i == m && j == n) {
            return 1;
        }
        return dfs(i + 1, j, m, n) + dfs(i, j + 1, m, n);
    }
    int uniquePaths(int m, int n) {
        return dfs(1, 1, m, n);
    }

    // 动态规划
    int uniquePaths1(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];
    }

    
};

63. 不同路径 II

力扣题目链接

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0
            return 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];
    }
};