Unique Path

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leetcode unique path

Unique Path

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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).

How many possible unique paths are there?

iamge
Above is a 3 x 7 grid. How many possible unique paths are there?

思路:这是一道明显的DP题，对于任意个点，它一定是又他的正上方的点或者左方的点移动过来的

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

Unique Path ii

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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.
与上面的题的唯一的区别是此时会有障碍物在移动的过程中

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        int dp[m][n];
        
        for(int i = 0; i < n; i++){
            if(obstacleGrid[0][i] == 0)
                dp[0][i] = 1;
            else{
                dp[0][i] = 0;
            }
        }
        
        for(int j = 0; j < m; j++){
            if(obstacleGrid[j][0] == 0)
                dp[j][0] = 1;
            else
                dp[j][0] = 0;
        }
            
        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++){
                if(obstacleGrid[i][j] == 0)
                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
                else
                    dp[i][j] = 0;
            }
            
        return dp[m - 1][n -1];
    }
};