Unique Paths II
題意：
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.
解題思路：
使用dp來作，時間複雜度為O(N^2)，空間複雜度為O(N^2)。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0) {
return 0;
}
int rows = obstacleGrid.length;
int cols = obstacleGrid[0].length;
int[][] res = new int[rows][cols];
if (obstacleGrid[0][0] == 1) {
res[0][0] = 0;
} else {
res[0][0] = 1;
}
for (int i = 1; i < cols; i++) {
if (obstacleGrid[0][i] != 1) {
res[0][i] = res[0][i - 1];
} else {
res[0][i] = 0;
}
}
for (int i = 1; i < rows; i++) {
if (obstacleGrid[i][0] != 1) {
res[i][0] = res[i - 1][0];
} else {
res[i][0] = 0;
}
}
for (int i = 1; i < rows; i++) {
for (int j = 1; j < cols; j++) {
if (obstacleGrid[i][j] == 0) {
res[i][j] = res[i - 1][j] + res[i][j - 1];
} else {
res[i][j] = 0;
}
}
}
return res[rows - 1][cols - 1];
}
}
上面的程式碼太雜了，有進步的空間，一開始先判斷開頭是否能走，再來用一維陣列來節省空間
程式碼如下：
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
//加了判斷a[0][0]的條件，如果一開始1的話，後面也不用算了
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1) {
return 0;
}
int rows = obstacleGrid.length;
int cols = obstacleGrid[0].length;
int[] res = new int[cols];
res[0] = 1;
for (int j = 1; j < cols; j++) {
if (obstacleGrid[0][j] == 1) {
res[j] = 0;
} else {
res[j] = res[j - 1];
}
}
for (int i = 1; i < rows; i++) {
// 判斷目前i,0是否有路走，如果有路走，則直接抄下res[0]，res[0]代表res[i - 1][0]
res[0] = obstacleGrid[i][0] == 1 ? 0 : res[0];
for (int j = 1; j < cols; j++) {
if (obstacleGrid[i][j] == 0) {
res[j] = res[j] + res[j - 1];
} else {
res[j] = 0;
}
}
}
return res[cols - 1];
}
}