[leetcode]63. Unique Paths II(Java)

https://leetcode.com/problems/unique-paths-ii/#/description

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.

package go.jacob.day628;public class Demo1 {/* * 把1换成0. * Runtime: 1 ms.Your runtime beats 20 % of java */public int uniquePathsWithObstacles(int[][] obstacleGrid) {if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0)return 0;int m = obstacleGrid.length;int n = obstacleGrid[0].length;//与1进行异或obstacleGrid[0][0] ^= 1;for (int i = 1; i < m; i++) {obstacleGrid[i][0] = obstacleGrid[i][0] == 1 ? 0 : obstacleGrid[i - 1][0];}for (int i = 1; i < n; i++) {obstacleGrid[0][i] = obstacleGrid[0][i] == 1 ? 0 : obstacleGrid[0][i - 1];}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {obstacleGrid[i][j] = obstacleGrid[i][j] == 1 ? 0 : obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];}}return obstacleGrid[m - 1][n - 1];}/* * Solution by me Runtime: 40 ms.Your runtime beats 0.45 % of java * submissions. */public int uniquePathsWithObstacles_1(int[][] obstacleGrid) {if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0)return 0;int m = obstacleGrid.length;int n = obstacleGrid[0].length;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (obstacleGrid[i][j] == 1)obstacleGrid[i][j] = -1;}}for (int i = 0; i < m; i++) {if (i == 0) {if (obstacleGrid[i][0] != -1)obstacleGrid[i][0] = 1;} else {if (obstacleGrid[i - 1][0] == -1 || obstacleGrid[i][0] == -1)obstacleGrid[i][0] = -1;elseobstacleGrid[i][0] = 1;}}for (int i = 0; i < n; i++) {if (i == 0) {if (obstacleGrid[0][i] != -1)obstacleGrid[0][i] = 1;} else {if (obstacleGrid[0][i - 1] == -1 || obstacleGrid[0][i] == -1)obstacleGrid[0][i] = -1;elseobstacleGrid[0][i] = 1;}}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {if (obstacleGrid[i][j] == -1) {continue;}if (obstacleGrid[i - 1][j] != -1 && obstacleGrid[i][j - 1] != -1)obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];else if (obstacleGrid[i - 1][j] == -1)obstacleGrid[i][j] = obstacleGrid[i][j - 1];elseobstacleGrid[i][j] = obstacleGrid[i - 1][j];}}return obstacleGrid[m - 1][n - 1] == -1 ? 0 : obstacleGrid[m - 1][n - 1];}}