leetcode 62. Unique Paths 、63. Unique Paths II

62. Unique Paths

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

63. Unique Paths II

leetcode的例子中int会越界，所以需要用long

与Unique Paths I不同在于多了障碍物，障碍物的情况直接为0就好，在初始化的时候需要做这个操作，在dp的迭代过程中也要做，其他与Unique Paths I 是一样的

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        if(m <= )
            return ;
        int n = obstacleGrid[].size();
        if(n <= )
            return ;
        if(obstacleGrid[][] ==  || obstacleGrid[m-][n-] == )
            return ;
        vector<vector<long> > dp(m,vector<long>(n));
        dp[][] = ;
        for(int i = ;i < m;i++){
            if(dp[i-][] ==  || obstacleGrid[i][] == )
                dp[i][] = ;
            else
                dp[i][] = ;
        }
        for(int i = ;i < n;i++){
            if(dp[][i-] ==  || obstacleGrid[][i] == )
                dp[][i] = ;
            else
                dp[][i] = ;
        }
        for(int i = ;i < m;i++){
            for(int j = ;j < n;j++){
                if(obstacleGrid[i][j] == )
                    dp[i][j] = ;
                else
                    dp[i][j] = dp[i-][j] + dp[i][j-];
            }
        }
        return dp[m-][n-];
    }
};