Unique Paths II
2017-08-24 22:30
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题目
Follow up for "Unique Paths":
Now consider if some obstacles are added tothe grids. How many unique paths would there be?
An obstacle and empty space is markedas1and0respectively in the grid.
For example,
There is one obstacle in the middle of a3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is2.
Note: m and n will be at most 100.
实现
public class UniquePathTwo {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;// 行
int n = obstacleGrid[0].length;// 列
if(m == 0 || n == 0) {
return0;
}
if(obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1) {
return0;
}
int[][] dp = new int[m]
;
dp[0][0]= 1;
for(int i = 1; i < n; i++) {
if(obstacleGrid[0][i] == 1) {
dp[0][i]= 0;
}else {
dp[0][i]= dp[0][i - 1];
}
}
for(int i = 1; i < m; i++) {
if(obstacleGrid[i][0] == 1) {
dp[i][0]= 0;
}else {
dp[i][0]= dp[i - 1][0];
}
}
for(int i = 1; i < m; i++) {
for(int j = 1; j < n; j++) {
if(obstacleGrid[i][j] == 1) {
dp[i][j]= 0;
}else {
dp[i][j]= dp[i - 1][j] + dp[i][j - 1];
}
}
}
returndp[m - 1][n - 1];
}
}
Follow up for "Unique Paths":
Now consider if some obstacles are added tothe grids. How many unique paths would there be?
An obstacle and empty space is markedas1and0respectively in the grid.
For example,
There is one obstacle in the middle of a3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is2.
Note: m and n will be at most 100.
实现
public class UniquePathTwo {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;// 行
int n = obstacleGrid[0].length;// 列
if(m == 0 || n == 0) {
return0;
}
if(obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1) {
return0;
}
int[][] dp = new int[m]
;
dp[0][0]= 1;
for(int i = 1; i < n; i++) {
if(obstacleGrid[0][i] == 1) {
dp[0][i]= 0;
}else {
dp[0][i]= dp[0][i - 1];
}
}
for(int i = 1; i < m; i++) {
if(obstacleGrid[i][0] == 1) {
dp[i][0]= 0;
}else {
dp[i][0]= dp[i - 1][0];
}
}
for(int i = 1; i < m; i++) {
for(int j = 1; j < n; j++) {
if(obstacleGrid[i][j] == 1) {
dp[i][j]= 0;
}else {
dp[i][j]= dp[i - 1][j] + dp[i][j - 1];
}
}
}
returndp[m - 1][n - 1];
}
}
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