LeetCode-63. 不同路径 II

最近英文版的访问特别慢，转战中文吧

和上一题一样，递归会超时

//63 不同路径2,递归解法
int uniquePaths2(vector<vector<int>>& obstacleGrid, int m,int n) {
    if ((m ==  && n == ) || (m ==  && n == ) || (m ==  && n == )) {
        if (obstacleGrid[obstacleGrid.size() - ][obstacleGrid[].size() - ] == )
            return ;
        else
            return ;
    }
    int p1 = ;
    int p2 = ;
    //获得向下走的总数,当下面的数为1或者是到了最底层,向下的总数为0
    if (m ==  || obstacleGrid[m-][n-] ==  )
        p1 = ;
    else
        p1 = uniquePaths2(obstacleGrid,m-,n);
    //向右
    if (n ==  || obstacleGrid[m - ][n - ] ==  )
        p2 = ;
    else
        p2 = uniquePaths2(obstacleGrid, m, n - );
    return p1 + p2;
}
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    if (obstacleGrid[][] == )
        return ;
    return uniquePaths2(obstacleGrid,obstacleGrid.size(),obstacleGrid[].size());
}

使用数组可以通过，和上题相比，这题思路就更直接，但是效率比之前更低一点，使用的result和题目所给的格子大小一样，从右下向左上遍历，更加清晰，就是需要额外处理终点

//非递归解法
int uniquePathsWithObstacles2(vector<vector<int>>& obstacleGrid) {
    /*if (obstacleGrid[0][0] == 1 || obstacleGrid[obstacleGrid.size() - 1][obstacleGrid[0].size() - 1])
        return 0;*/
    vector<vector<int>> result(obstacleGrid.size(),vector<int>(obstacleGrid[].size(),));
    for (int i = obstacleGrid.size() - ; i >= ; i--) {
        for (int j = obstacleGrid[].size() - ; j >= ; j--) {
            if (obstacleGrid[i][j] == ) {
                result[i][j] = ;
                continue;
            }
            if (i == obstacleGrid.size() -  && j == obstacleGrid[].size() - ) {
                result[i][j] = ;
                continue;
            }
            int p1 = ;
            int p2 = ;
            //如果是下部有位置
            if (i != obstacleGrid.size() - )
                p1 = result[i + ][j];
            //如果是右边有位置
            if (j != obstacleGrid[].size() - )
                p2 = result[i][j + ];
            result[i][j] = p1 + p2 ;
        }
    }
    return result[][];
}