leetcode 63. Unique Paths II

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本文介绍了解决LeetCode 63题的独特路径问题的方法，通过动态规划算法计算在一个存在障碍物的网格中从左上角到右下角的不同路径数量。给出的代码实现了对网格中每个位置进行状态转移，最终得到所有可能路径的数量。

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/*
leetcode 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.

解题思路：动态规划

dp[j] = dp[j] + dp[j-1]

*/

#include <iostream>
#include <vector>
#include <string>

using namespace std;

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

        dp[0] = 1;      //第0行0列的步数

        for (int i = 0; i < m; ++i)
        {
            if (obstacleGrid[i][0] == 1)        //边界处理：有一个为1，说明此路不通
                dp[0] = 0;
            for (int j = 1; j < n; ++j)
            {
                //dp[j]为第i行，第j列的步数
                if (obstacleGrid[i][j] == 1)
                    dp[j] = 0;
                else
                    dp[j] = dp[j] + dp[j - 1];
                cout << i << "," << j << ":" << dp[j] << endl;
            }
        }

        return dp[n - 1];
    }
};

void test()
{
    vector<vector<int>> ob{
        {0,0,0},
        {0,1,0},
        {0,0,0}
    };

    Solution sol;
    cout << sol.uniquePathsWithObstacles(ob) << endl;
}

int main()
{
    test();

    return 0;
}