Unique Path II
2015-07-29 16:14
429 查看
同Leetcode No.64 Minimum Path一样,可以使用滚动矩阵,来减小Space Complexity。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid.length == 0) {
return -1;
}
int row = obstacleGrid.length;
int col = obstacleGrid[0].length;
int[] path = new int[col];
for (int j = 0; j < col; j++) {
if (obstacleGrid[0][j] == 0) {
path[j] = 1;
} else {
break;
}
}
for (int i = 1; i < row; i++) {
if (obstacleGrid[i][0] == 1) {
path[0] = 0;
}
for (int j = 1; j < col; j++) {
if (obstacleGrid[i][j] == 0) {
path[j] += path[j - 1];
} else {
path[j] = 0;
}
}
}
return path[col - 1];
}
}
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid.length == 0) {
return -1;
}
int row = obstacleGrid.length;
int col = obstacleGrid[0].length;
int[] path = new int[col];
for (int j = 0; j < col; j++) {
if (obstacleGrid[0][j] == 0) {
path[j] = 1;
} else {
break;
}
}
for (int i = 1; i < row; i++) {
if (obstacleGrid[i][0] == 1) {
path[0] = 0;
}
for (int j = 1; j < col; j++) {
if (obstacleGrid[i][j] == 0) {
path[j] += path[j - 1];
} else {
path[j] = 0;
}
}
}
return path[col - 1];
}
}
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