代码随想录第36天|动态规划

62. 不同路径

补充: 对二维数组的操作

dp[j][i] 表示到 j,i 有多少种路径
递推公式: dp[j][i] = dp[j - 1][i] + dp[j][i - 1]
初始化: dp[0][i] 和 dp[j][0] 都只有1种情况
遍历顺序: 由于dp[j][i] 由上和左的元素推导, 所以采用从左到右、从上到下的遍历顺序

class Solution {
public:
    void uniquePaths(int m, int n) {
        vector<vector<int>> dp(m, vector<int>(n, 0));
        //初始化
        for (int i = 0; i < dp[0].size(); i++) dp[0][i] = 1;
        for (int i = 0; i < dp.size(); i++) dp[i][0] = 1;
        //递推公式
        for (int j = 1; j < m; j++) {
            for (int i = 1; i < n; i++) {
                dp[j][i] = dp[j][i - 1] + dp[j - 1][i];
            }
        }
        //输出dp[j][i]
        for (int j = 0; j < dp.size(); j++) {
            for (int i = 0; i < dp[i].size(); i++) {
                cout << dp[j][i] << " ";
            }
            cout << endl;
        }
        return dp[m - 1][n - 1];                  
    }
};

63. 不同路径 II

参考

dp[j][i] 表示到(j, i) 后的路径数
递推公式
初始化 dp[0][i] 和 dp[j][0] 为 1 ,当遇到障碍物后变 0
遍历顺序同上

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int n = obstacleGrid[0].size();
        int m = obstacleGrid.size();        
        vector<vector<int>> dp = vector<vector<int>>(m, vector<int>(n, 0));
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 1) break;
            dp[0][i] = 1;
        }
        for (int j = 0; j < m; j++) {
            if (obstacleGrid[j][0] == 1) break;
            dp[j][0] = 1;
        }

        for (int j = 1; j < m; j++) {
            for (int i = 1; i < n; i++) {
                if (obstacleGrid[j][i] == 1) {//该点有障碍
                    dp[j][i] = 0;                    
                } else {
                    dp[j][i] = dp[j][i - 1] + dp[j - 1][i];
                }              
            }
        }
        return dp[m - 1][n - 1];
    }
};