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[LeetCode]Unique Paths II

2015-04-06 16:43 218 查看
该题和之前的题目之间的区别就是某些单元是不可达的,这可以加上判断语句即可

算法1:

public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int pathsNum = 0,m = obstacleGrid.length,n=obstacleGrid[0].length;
if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1)return 0;

int [][] paths = new int[m]
;
for(int i = 0; i< m; i ++){
for(int j = 0; j < n; j ++){
if(i > 0 && j > 0){
if(obstacleGrid[i][j-1] == 1 &&  obstacleGrid[i-1][j] ==0){
paths[i][j] = paths[i-1][j];
}else if(obstacleGrid[i-1][j] == 1 && obstacleGrid[i][j-1] == 0){
paths[i][j] = paths[i][j-1];
}else if(obstacleGrid[i-1][j] == 0 && obstacleGrid[i][j-1] == 0){
paths[i][j] = paths[i][j-1] + paths[i-1][j];
}else {
paths[i][j] = 0;
}
}else if(i > 0 && j == 0){
if(obstacleGrid[i-1][j] == 0){
paths[i][j] = paths[i-1][j];
}else {
paths[i][j] = 0;
}

}else if(i == 0 && j > 0){
if(obstacleGrid[i][j-1] == 0){
paths[i][j] = paths[i][j-1];
}else {
paths[i][j] = 0;
}
}else {
if(obstacleGrid[0][0] == 1)
return 0;
else paths[0][0] = 1;
}
}
}

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

}
感觉自己做的太复杂了,别人的简单版本如下:

public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if(obstacleGrid == null || obstacleGrid.length==0 || obstacleGrid[0].length==0)
return 0;
int[] res = new int[obstacleGrid[0].length];
res[0] = 1;
for(int i=0;i<obstacleGrid.length;i++)
{
for(int j=0;j<obstacleGrid[0].length;j++)
{
if(obstacleGrid[i][j]==1)
{
res[j]=0;
}
else
{
if(j>0)
res[j] += res[j-1];
}
}
}
return res[obstacleGrid[0].length-1];
}
}
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