[LeetCode] 63.Unique Paths II

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).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

62. Unique Paths 的拓展， 有如下不一樣：

1. 當(i, j)有障礙時dp[i][j] = 0

2. dp[0][j]和dp[i][0]未必爲1.

dp[0][j] = obstacleGrid[0][j] ? 0 : dp[0][j-1]

dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0]

3. 當obstacleGrid [0][0] = 1時，return 0

解法：DP, 創建二維數組，dp[i][j]表示在某一位置能到達的不一樣路徑數量，dp[i][j] = dp[i-1][j] + dp[i][j-1] ，若是某一位置grid[i][j]=1說明爲障礙物，那麼dp[i][j] = 0，還要注意i, j爲0的狀況。

 

Java: 

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) { 
        int m = obstacleGrid.length; 
        int n = obstacleGrid[0].length; 

        if (m == 0 || n == 0) { 
            return 0;
        }

        if (obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) {
            return 0;
        }

        int[][] dp = new int[m][n];

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

Java:

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    int width = obstacleGrid[0].length;
    int[] dp = new int[width];
    dp[0] = 1;
    for (int[] row : obstacleGrid) {
        for (int j = 0; j < width; j++) {
            if (row[j] == 1)
                dp[j] = 0;
            else if (j > 0)
                dp[j] += dp[j - 1];
        }
    }
    return dp[width - 1];
}　　

Python:

class Solution:

    def uniquePathsWithObstacles(self, obstacleGrid):
        mp = obstacleGrid
        for i in range(len(mp)):
            for j in range(len(mp[i])):
                if i == 0 and j == 0:
                    mp[i][j] = 1 - mp[i][j]
                elif i == 0:
                    if mp[i][j] == 1:
                        mp[i][j] = 0
                    else:
                        mp[i][j] = mp[i][j - 1]
                elif j == 0:
                    if mp[i][j] == 1:
                        mp[i][j] = 0
                    else:
                        mp[i][j] = mp[i - 1][j]
                else:
                    if mp[i][j] == 1:
                        mp[i][j] = 0
                    else:
                        mp[i][j] = mp[i - 1][j] + mp[i][j - 1]
        if mp[-1][-1] > 2147483647: 
            return -1
        else:
            return mp[-1][-1]

Python: wo

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for i in xrange(m)]
        for i in xrange(m):
            for j in xrange(n):
                if obstacleGrid[i][j] == 1:
                    dp[i][j] == 0
                else:
                    if i == 0 and j == 0:  # 寫錯成了 or 
                        dp[i][j] = 1
                    elif i == 0:
                        dp[i][j] = dp[i][j-1]
                    elif j == 0:
                        dp[i][j] = dp[i-1][j]                        
                    else:
                        dp[i][j] = dp[i-1][j] + dp[i][j-1]
                        
        return dp[-1][-1]

Python: wo

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for i in xrange(m)]
        for i in xrange(m):
            for j in xrange(n):
                if obstacleGrid[i][j] != 1:
                    if i == 0 and j == 0:
                        dp[i][j] = 1
                    elif i == 0:
                        dp[i][j] = dp[i][j-1]
                    elif j == 0:
                        dp[i][j] = dp[i-1][j]                        
                    else:
                        dp[i][j] = dp[i-1][j] + dp[i][j-1]
                        
        return dp[-1][-1]　　

C++:

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

　　

　

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[LeetCode] 62.Unique Paths