62. Unique Paths

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

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

class Solution {
public:
    int uniquePaths(int m, int n) {
        int dp[m][n] = {0};

        for(int i = 0; i < n; i++)
            dp[0][i] = 1;
        for(int j = 0; j < m; j++)
            dp[j][0] = 1;

        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++){
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }

        return dp[m - 1][n -1];
    }
};

状态压缩后的dp

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m == 0 || n == 0)    return 0;

        vector<int> dp(m, 1);

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

unique path 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.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        if(m == 0 || n == 0)    return 0;

        vector<int> dp(m, 1);


        dp[0] = 1 - obstacleGrid[0][0];

        //初始化第0列
        for(int i = 1; i < m; i++){
            if(obstacleGrid[i][0] == 0 && dp[i - 1])
                dp[i] = 1;
            else
                dp[i] = 0;
        }

        for(int i = 1; i < n; i++){
            dp[0] = (dp[0] &&obstacleGrid[0][i] == 0)?1:0;
            std::cout<<dp[0]<<std::endl;
            for(int j = 1; j < m; j++){
                if(obstacleGrid[j][i] == 0)
                    dp[j] = dp[j - 1] + dp[j];
                else
                    dp[j] = 0;
            }
        }
        return dp[m - 1];
    }
};