Leetcode:63. Unique Paths II

Description

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.

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.

思路

• 同前一个题一样的动态规划，用一个二维数组记录状态,dp[i][j]表示从(0,0)到(i,j)的路径数
• dp[i][j] = dp[i][j-1] + dp[i-1][j]，若obstacleGrid[i][j] = 1, dp[i][j] = 0
• 可以把二维数组降到一维

代码

• 二维数组

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

• 一维数组

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