63. Unique Paths II

题目：

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.

Note: m and n will be at most 100.

Hide Tags

 Array Dynamic Programming

 

链接：  http://leetcode.com/problems/unique-paths-ii/

题解：

也是DP问题，Unique Path一样可以in place解决。要点是在设置第一行和第一列碰到obstacle的时候，要将其以及之后的所有值设置为零，因为没有路径可以达到。之后在DP扫描矩阵的时候，也要讲obstacle所在的位置清零。

Time Complexity - O(m * n)， Space Complexity - O(1)。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int rowNum = obstacleGrid.length, colNum = obstacleGrid[0].length;
        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1){
            return 0;
        }
        
        for(int row = 0; row < rowNum; row ++){
            if(obstacleGrid[row][0] == 0)
                obstacleGrid[row][0] = 1;
            else if (obstacleGrid[row][0] == 1){              //if find obstacle, set all [row,0] below obstacle to 0
                for(int tempRow = row; tempRow < rowNum; tempRow ++)
                    obstacleGrid[tempRow][0] = 0;
                break;
            }
        }
        
         for(int col = 1; col < colNum; col ++){
            if(obstacleGrid[0][col] == 0)
                obstacleGrid[0][col] = 1;
            else if (obstacleGrid[0][col] == 1){            // //if find obstacle, set all [0,col] one the right of obstacle to 0
                for(int tempCol = col; tempCol < colNum; tempCol ++)
                    obstacleGrid[0][tempCol] = 0;
                break;
            }
        }
        
        for(int i = 1; i < rowNum; i ++){
            for(int j = 1; j < colNum; j ++){
                if(obstacleGrid[i][j] == 1)
                    obstacleGrid[i][j] = 0;
                else
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
            }
        }
        
        return obstacleGrid[rowNum - 1][colNum - 1];
    }
}

二刷：

和一刷一样， 就是先判断行和列中的obstacle元素，将其与其之后的为止置零。接下来遍历整个矩阵。

Java:

Time Complexity - O(m * n)， Space Complexity - O(1)。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1) {
            return 0;
        }
        int rowNum = obstacleGrid.length, colNum = obstacleGrid[0].length;
        for (int i = 0; i < rowNum; i++) {
            if (obstacleGrid[i][0] == 1) {
                for (int k = i; k < rowNum; k++) {
                    obstacleGrid[k][0] = 0;
                }
                break;
            } else {
                obstacleGrid[i][0] = 1;
            }
        }    
        for (int j = 1; j < colNum; j++) {
            if (obstacleGrid[0][j] == 1) {
                for (int k = j; k < colNum; k++) {
                    obstacleGrid[0][k] = 0;
                }
                break;
            } else {
                obstacleGrid[0][j] = 1;
            }
        }    
        for (int i = 1; i < rowNum; i++) {
            for (int j = 1; j < colNum; j++) {
                if (obstacleGrid[i][j] == 1) {
                    obstacleGrid[i][j] = 0;
                } else {
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
                }
            }
        }
        return obstacleGrid[rowNum - 1][colNum - 1];
    }
}

三刷:

Java:

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1) {
            return 0;
        }
        int rowNum = obstacleGrid.length, colNum = obstacleGrid[0].length;
        for (int i = 0; i < rowNum; i++) {
            if (obstacleGrid[i][0] == 1) {
                for (int k = i; k < rowNum; k++) {
                    obstacleGrid[k][0] = 0;
                }
                break;
            } else {
                obstacleGrid[i][0] = 1;
            }
        }    
        for (int j = 1; j < colNum; j++) {
            if (obstacleGrid[0][j] == 1) {
                for (int k = j; k < colNum; k++) {
                    obstacleGrid[0][k] = 0;
                }
                break;
            } else {
                obstacleGrid[0][j] = 1;
            }
        }    
        for (int i = 1; i < rowNum; i++) {
            for (int j = 1; j < colNum; j++) {
                obstacleGrid[i][j] = obstacleGrid[i][j] == 1 ? 0 : obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
            }
        }
        return obstacleGrid[rowNum - 1][colNum - 1];
    }
}