题目

解答

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;if (obstacleGrid[0][0] == 1) {return 0;}if (obstacleGrid[m - 1][n - 1] == 1) {return 0;}int[][] dp = new int[m][n];dp[0][0] = 1;for (int i = 1, imax = m; i < imax; ++i) {if (obstacleGrid[i][0] == 1) {dp[i][0] = 0;} else {dp[i][0] = dp[i - 1][0];}}for (int j = 1, jmax = n; j < jmax; ++j) {if (obstacleGrid[0][j] == 1) {dp[0][j] = 0;} else {dp[0][j] = dp[0][j - 1];}}for (int i = 1, imax = m; i < imax; ++i) {for (int j = 1, jmax = n; j < jmax; ++j) {if (obstacleGrid[i][j] == 1) {dp[i][j] = 0;} else {if (obstacleGrid[i - 1][j] == 0) {dp[i][j] += dp[i - 1][j];}if (obstacleGrid[i][j - 1] == 0) {dp[i][j] += dp[i][j - 1];}}}}return dp[m - 1][n - 1];}
}

要点
本题目充分说明，使用动态规划解题时，初始值很重要。
另外，假如起点和终点均为障碍物的话，可以直接返回，不需要执行后续的求解操作。

准备的用例，如下

@Before
public void before() {t = new Solution();
}@Test
public void test001() {assertEquals(2, t.uniquePathsWithObstacles(new int[][] { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } }));
}@Test
public void test002() {assertEquals(1, t.uniquePathsWithObstacles(new int[][] { { 0, 1 }, { 0, 0 } }));
}@Test
public void test003() {assertEquals(1, t.uniquePathsWithObstacles(new int[][] { { 0, 0 } }));
}@Test
public void test004() {assertEquals(0, t.uniquePathsWithObstacles(new int[][] { { 0, 0 }, { 1, 1 }, { 0, 0 } }));
}@Test
public void test005() {assertEquals(0, t.uniquePathsWithObstacles(new int[][] { { 0, 0 }, { 0, 1 } }));
}@Test
public void test006() {assertEquals(0, t.uniquePathsWithObstacles(new int[][] { { 1, 0 }, { 0, 0 } }));
}@Test
public void test007() {assertEquals(0, t.uniquePathsWithObstacles(new int[][] { { 0, 1, 0, 0, 0 }, { 1, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0 } }));
}