《LeetCode之每日一题》:26. 不同路径 II

Posted 2021-05-16 是七喜呀！

不同路径 II

• 有关题目
• • 题解

题目链接： 不同路径 II

有关题目

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？

提示：

m == obstacleGrid.length
n == obstacleGrid[i].length
1 <= m, n <= 100
obstacleGrid[i][j] 为 0 或 1

题解

1、动态规划
//这边目前在原数组基础上改数据的情况不知道怎么解决

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

该动态规划通过

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));
    for (int i = 0; i < m; i++)
    {
        dp[i][0] = (obstacleGrid[i][0] == 1 ? 0 : 1);
        if (obstacleGrid[i][0] == 1)
            break;
    }
    for (int j = 0; j < n; j++)
    {
        dp[0][j] = (obstacleGrid[0][j] == 1 ? 0 : 1);
        if (obstacleGrid[0][j] == 1)
            break;
    }
    for (int i = 1; i < m; i++)
    {
        for (int j = 1; j < n; j++)
        {
            dp[i][j] =(obstacleGrid[i][j] == 0) ? (dp[i - 1][j] + dp[i][j - 1]):0;
        }
    }
    return dp[m - 1][n - 1];
    }
};

时间复杂度：O(mn)
空间复杂度：O(mn)

2、滑动数组

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    int m = obstacleGrid.size();
    int n = obstacleGrid[0].size();
    vector<int> dp(n);

    dp[0] = (obstacleGrid[0][0] == 0);
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (obstacleGrid[i][j] == 1)
            {
                dp[j] = 0;
                continue;
            }
            if (j - 1 >= 0)
                dp[j] += dp[j - 1];
        }
    }
    return dp.back();
    }
};

时间复杂度：O(mn)
空间复杂度：O(m)

C版本

int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize) {
    int n = obstacleGridSize, m = obstacleGridColSize[0];
    int f[m];
    memset(f, 0, sizeof(f));
    f[0] = (obstacleGrid[0][0] == 0);
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (obstacleGrid[i][j] == 1) {
                f[j] = 0;
                continue;
            }
            if (j - 1 >= 0 && obstacleGrid[i][j - 1] == 0) {
                f[j] += f[j - 1];
            }
        }
    }
    return f[m - 1];
}