LeetCode_63---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.

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 Array Dynamic Programming

翻译：

Code:

package From61;import java.util.Arrays;/** * @author MohnSnow * @time 2015年6月29日 上午9:43:51 *  */public class LeetCode63 {/** * @param argsmengdx *            -fnst *///核心地方在于一个点只能有两个方向过来，只是在上一个题的基础上演变过来。//296msAC---http://www.cnblogs.com/chkkch/archive/2012/11/15/2772282.htmlpublic static int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;System.out.println("----------old---------------");for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {System.out.print(" " + obstacleGrid[i][j]);}System.out.println("");}System.out.println("----------new---------------");if (m <= 0 || n <= 0 || obstacleGrid[0][0] == 1) {return 0;}int temp[][] = new int[m][n];temp[0][0] = 1;for (int i = 1; i < n; i++) {if (obstacleGrid[0][i] == 0) {temp[0][i] = temp[0][i - 1];} else {temp[0][i] = 0;}}for (int i = 1; i < m; i++) {if (obstacleGrid[i][0] == 0) {temp[i][0] = temp[i - 1][0];} else {temp[i][0] = 0;}}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {if (obstacleGrid[i][j] == 0) {temp[i][j] = temp[i - 1][j] + temp[i][j - 1];} else {temp[i][j] = 0;}}}for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {System.out.print(" " + temp[i][j]);}System.out.println("");}return temp[m - 1][n - 1];}public static void main(String[] args) {int[][] obstacleGrid = {{ 0, 0, 0, 0, 1 },{ 0, 0, 0, 0, 0 },{ 0, 0, 0, 0, 0 },{ 0, 1, 0, 1, 1 },{ 0, 0, 0, 0, 0 },{ 0, 0, 0, 0, 0 } };System.out.println("uniquePaths: " + uniquePathsWithObstacles(obstacleGrid));}}