leetcode 62|63. Unique Paths 1|2_leetcode uniqu path-优快云博客

本文链接：https://blog.youkuaiyun.com/liqinzhe11/article/details/78410741

本文介绍了解决独特路径问题的两种动态规划方法：一种是在没有障碍的情况下寻找从左上角到右下角的所有可能路径数量；另一种则考虑了网格中存在障碍的情况。通过详细解释算法逻辑并给出具体实现代码，帮助读者理解如何高效地解决这两类问题。

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62. Unique Paths

标准动态规划，当前步只能从上面和左边走过来。

class Solution {
public:
    int uniquePaths(int m, int n) 
    {
        vector<vector<int>> ret(n, vector<int> (m, 0));
        for (int i = 0; i < m; i++)
             ret[0][i] = 1;
        for (int i = 0; i < n; i++)
             ret[i][0] = 1;
        for (int i = 1; i < n; i++)
        {
            for(int j = 1; j < m; j++)
            {
                ret[i][j] = ret[i-1][j] + ret[i][j-1];
            }
        }  
        return ret[n-1][m-1];  
    }
};

63. Unique Paths II

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

还是动态规划，加一个条件而已

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) 
    {
        int row = obstacleGrid.size();
        int col = obstacleGrid[0].size();
        if (row > 0 && col > 0 && obstacleGrid[0][0] == 1)
            return 0;
        
        int flag = 1;
        for (int i = 0; i < row; i++)
        {
            if (flag == 1 && obstacleGrid[i][0] == 0) 
                obstacleGrid[i][0] = 1; 
            else
            {
                flag = 0;
                obstacleGrid[i][0] = 0; 
            }
        }
        flag = 1;
        for(int i = 1; i < col; i++)
        {
            if (flag == 1 && obstacleGrid[0][i] == 0) 
                obstacleGrid[0][i] = 1; 
            else
            {
                flag = 0;
                obstacleGrid[0][i] = 0; 
            }
        }
         
        for(int i = 1; i < row; i++)
            for(int j = 1; j < col; j++)
            {
                 if (obstacleGrid[i][j] == 1)
                     obstacleGrid[i][j] = 0;
                 else
                     obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
            }
        return obstacleGrid[row-1][col-1];
    }
};