C++ 解法, 执行用时: 4ms, 内存消耗: 420KB, 提交时间: 2022-06-17

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param obstacleGrid int整型vector<vector<>> 
     * @return int整型
     */
    int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {
        // write code here
        if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0){
            return 0;
        }
        vector<int> dp = obstacleGrid[0];
        for(int i = 1; i < obstacleGrid.size(); i++){
            int pv = obstacleGrid[i][0];
            int accum = pv;
            for(int j = 1; j < obstacleGrid[0].size(); j++){
                dp[j] = obstacleGrid[i][j] * (pv + dp[j]);
                pv = dp[j];
                accum += pv;
            }
            if (accum == 0){
                return 0;
            }
        }
        return dp.back();
    }
};

C++ 解法, 执行用时: 4ms, 内存消耗: 512KB, 提交时间: 2022-04-02

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param obstacleGrid int整型vector<vector<>> 
     * @return int整型
     */
    int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {
        // write code here
        int size_row =obstacleGrid.size();
        int size_col = obstacleGrid[0].size();
        vector< vector<int> > dp(size_row+1,vector<int>(size_col+1,0));
        for(int i=0;i<size_row;i++) {
            for(int j = 0; j< size_col; j++) {
                
                //dp[i][j] =dp[i-1][j]+dp[i][j-1];
                if((i == 0 || j==0) && obstacleGrid[i][j] ==1) dp[i][j] =1;
                else {
                    if(obstacleGrid[i][j] ==1 ) dp[i][j] = dp[i-1][j]+dp[i][j-1];
                }
                
            }
        }
        return dp[size_row-1][size_col-1];
    }
};

C++ 解法, 执行用时: 4ms, 内存消耗: 520KB, 提交时间: 2022-03-04

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param obstacleGrid int整型vector<vector<>> 
     * @return int整型
     */
    int uniquePathsWithObstacles(vector<vector<int> >& obs) {
        // write code here
        int m = obs.size();
        int n = obs[0].size();
        vector<vector<int> > route(m,vector<int>(n,1));
        route[0][0] = obs[0][0]==1?1:0;
        for(int i = 1;i<m;++i){
            for(int j = 1;j<n;++j){
                if(obs[i][j - 1] == 1 && obs[i-1][j] == 1){
                    route[i][j]=route[i][j-1] + route[i-1][j];
                } else if(obs[i][j - 1] == 0 && obs[i-1][j] == 1){
                    route[i][j] = route[i-1][j];
                } else if(obs[i][j - 1] == 1 && obs[i-1][j] == 0){
                    route[i][j] = route[i][j-1];
                } else {
                    route[i][j] = 0;
                }
            }
        }
        return route[m-1][n-1];
    }
};

C++ 解法, 执行用时: 4ms, 内存消耗: 524KB, 提交时间: 2022-07-20

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param obstacleGrid int整型vector<vector<>> 
     * @return int整型
     */
    int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {
        // write code here
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<int> dp(n, 0);
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 0) {
                dp[i] = 0;
            }
            else if (i == 0) {
                dp[i] = 1;
            }
            else {
                dp[i] = dp[i - 1];
            }
        }

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

C++ 解法, 执行用时: 4ms, 内存消耗: 524KB, 提交时间: 2022-06-22

class Solution {
  public:
    int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {
        if (obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0) {return 0;}       
        vector<int> a = obstacleGrid[0];        
        for (int i = 1; i < obstacleGrid.size(); i++) {
            int b = obstacleGrid[i][0];
            int c = b;
            for (int j = 1; j < obstacleGrid[0].size(); j++) {
                a[j] = obstacleGrid[i][j] * (b + a[j]);
                b = a[j];
                c += b;
            }
            if (c == 0) {return 0;}
        }        
        return a.back();
    }
};