Unique Paths II Solutions in C++

// Source : https://oj.leetcode.com/problems/unique-paths-ii/
// Author : Hao Chen
// Date   : 2014-06-25



#include <iostream>
#include <vector>
using namespace std;

//As same as DP solution with "Unique Path I", just need to consider the obstacles.
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    vector<vector<unsigned int>>  v (row, vector<unsigned int>(col, 0));
    unsigned int max=0;
    for (int i=0; i<obstacleGrid.size(); i++){
        for (int j=0; j<obstacleGrid[0].size(); j++){
            if(obstacleGrid[i][j] == 1){
                max = v[i][j] = 0;
            } else {
                if (i>0 && j>0) {
                    max= v[i][j] = v[i-1][j] + v[i][j-1];
                }else if(i>0){
                    max = v[i][j] = v[i-1][j];
                }else if(j>0){
                    max = v[i][j] = v[i][j-1];
                }else{
                    max = v[i][j] = 1 ;
                }
            }
        }
    }
    return max;
}

// the previous implemetation has too many if-else
// the following dynamic programming is much more easy to read
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    int row = obstacleGrid.size();
    int col = obstacleGrid[0].size();

    vector< vector <unsigned int> >  dp (row, vector<unsigned int>(col, 0));

    dp[0][0] =  obstacleGrid[0][0]  ?  0 : 1;
    for (int r=1; r<row; r++) {
        dp[r][0] = obstacleGrid[r][0]  ?  0 : dp[r-1][0];
    }
    for (int c=1; c<col; c++) {
        dp[0][c] = obstacleGrid[0][c]  ?  0 : dp[0][c-1];
    }

    for (int r=1; r<row; r++) {
        for (int c=1; c<col; c++) {
            dp[r][c]  = obstacleGrid[r][c] == 1 ? 0 : dp[r][c-1] + dp[r-1][c];
        }
    }

    return dp[row-1][col-1];
}

#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>
using std::vector; using std::prev;

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        if (obstacleGrid.empty()) return 0;
        vector<int> vec(obstacleGrid.front().cbegin(), obstacleGrid.front().cend());
        auto flag = std::find(vec.begin(), vec.end(), 1);
        std::fill(vec.begin(), flag, 1);
        std::fill(flag, vec.end(), 0);
        for (auto line = std::next(obstacleGrid.cbegin()); line != obstacleGrid.cend(); ++line) {
            auto iter = vec.begin();
            for (auto i : *line) {
                if (i) *iter = 0;
                else if (iter != vec.begin()) *iter += *prev(iter);
                ++iter;
            }
        }   
        return vec.back();
    }
};

/// Source : https://leetcode.com/problems/unique-paths-ii/
/// Author : liuyubobobo
/// Time   : 2018-10-25
/// Updated: 2019-09-24

#include <iostream>
#include <vector>

using namespace std;


/// Dynamic Programming
/// Time Complexity: O(m*n)
/// Space Complexity: O(m*n)
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {

        int m = obstacleGrid.size();
        if(!m) return 0;

        int n = obstacleGrid[0].size();
        if(!n || obstacleGrid[0][0])
            return 0;

        vector<vector<long long>> dp(m, vector<long long>(n, 1ll));
        dp[0][0] = 1;
        for(int j = 1; j < n; j ++)
            if(obstacleGrid[0][j])
                dp[0][j] = 0;
            else
                dp[0][j] = dp[0][j - 1];

        for(int i = 1; i < m; i ++)
            if(obstacleGrid[i][0])
                dp[i][0] = 0;
            else
                dp[i][0] = dp[i - 1][0];

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


int main() {

    return 0;
}