代码随想录算法训练营第31天 | 第九章动态规划02

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今日记录

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62.不同路径

Leetcode链接
1.确定dp数组（dp table）以及下标的含义：dp[i][j] 含义 到达 (i,j) 处有 dp[i][j] 种路径
2.确定递推公式：dp[i][j] = dp[i-1][j] + dp[i][j-1]
3.dp数组如何初始化: for i=0 ~ m-1, dp[i][0]=1, for j=0 ~ n-1, dp[0][j]=1
4.确定遍历顺序：从左向右，从上往下
5.举例推导dp数组

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];
    }
};

63.不同路径Ⅱ

Leetcode链接

1.确定dp数组（dp table）以及下标的含义: dp[i][j] 含义 到达 (i,j) 处有 dp[i][j] 种路径
2.确定递推公式: if (obstacleGrid[i][j] == 1 ) comtinue; dp[i][j] = dp[i-1][j] + dp[i][j-1]
3.dp数组如何初始化: for i=0 ~ m-1 且 obstacleGrid[i][0] == 0, dp[i][0]=1, for j=0 ~ n-1 且 obstacleGrid[0][j] == 0, dp[0][j]=1
4.确定遍历顺序
5.举例推导dp数组

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));
        // 初始化
        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;
        // 遍历循环
        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];
    }
};

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总结