Unique Paths II 求在矩阵里从左上走到右下的总方法数（有障碍）@LeetCode

package Level3;

/**
* Unique Paths II
*
* 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.
*/
public class S63 {

public static void main(String[] args) {

}

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

// DP数组用来存放到每一个格子时的路径数目
int[][] cnt = new int[m]
;
if(obstacleGrid[0][0] == 1){
cnt[0][0] = 0;
}else{
cnt[0][0] = 1;
}

// corner case 如果最后一个点是obstacle则没有办法到达
if(obstacleGrid[m-1][n-1] == 1){
return 0;
}

// 第一列的格子只有一种到达方式（向下）
for(int i=1; i<m; i++){
if(obstacleGrid[i][0]!=1 && cnt[i-1][0]!=0){
cnt[i][0] = 1;
}
}
// 第一行的格子只有一种到达方式（向右）
for(int i=1; i<n; i++){
if(obstacleGrid[0][i]!=1 && cnt[0][i-1]!=0){
cnt[0][i] = 1;
}
}

for(int i=1; i<m; i++){
for(int j=1; j<n; j++){
if(obstacleGrid[i-1][j] != 1){
cnt[i][j] += cnt[i-1][j];
}
if(obstacleGrid[i][j-1] != 1){
cnt[i][j] += cnt[i][j-1];
}
}
}

return cnt[m-1][n-1];
}

}

public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int[][] path = new int[obstacleGrid.length][obstacleGrid[0].length];
for(int i=0; i<obstacleGrid.length; i++) {
if(obstacleGrid[i][0] == 1) {
break;
}
path[i][0] = 1;
}
for(int j=0; j<obstacleGrid[0].length; j++) {
if(obstacleGrid[0][j] == 1) {
break;
}
path[0][j] = 1;
}

for(int i=1; i<obstacleGrid.length; i++) {
for(int j=1; j<obstacleGrid[0].length; j++) {
if(obstacleGrid[i][j] == 1) {
continue;
}
if(obstacleGrid[i-1][j] != 1) {
path[i][j] += path[i-1][j];
}
if(obstacleGrid[i][j-1] != 1) {
path[i][j] += path[i][j-1];
}
}
}

return path[obstacleGrid.length-1][obstacleGrid[0].length-1];
}
}