算法训练营第三十四天（8.18）| 动态规划Part02：不同路径

目录

Leecode 62.不同路径

 Leecode 63.不同路径 II

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Leecode 62.不同路径

题目地址：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

题目类型：基础动态规划

class Solution {
public:
    int uniquePaths(int m, int n) {
        // 代表到达第(i,j)个点的路径个数    (0,0)为开始点，路径数为1，且第一行和第一列均为1
        vector<vector<int>> dp(m, vector<int>(n, 1));
        // 状态转移方程为：dp[i][j] = dp[i - 1][j] + dp[i][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];
    }
};

 Leecode 63.不同路径 II

题目地址：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

题目类型：基础动态规划

 可以原地DP，从而空间优化

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid[0][0] == 1) return 0;
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        // 代表到达(i, j)的路径数目，若某一个点是障碍，则为0
        for (int i = 0; i < m; ++i) {
            if (obstacleGrid[i][0] == 1) {
                for (int j = i; j < m; ++j) obstacleGrid[j][0] = 0;
                break;
            }
            else obstacleGrid[i][0] = 1;
        }
        for (int i = 1; i < n; ++i) {
            if (obstacleGrid[0][i] == 1) {
                for (int j = i; j < n; ++j) obstacleGrid[0][j] = 0;
                break;
            }
            else obstacleGrid[0][i] = 1;
        }
        
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (obstacleGrid[i][j] == 1) obstacleGrid[i][j] = 0;
                else obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
            }
        }
        return obstacleGrid[m - 1][n - 1];
    }
};