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63. 不同路径 II
给定一个 m x n 的整数数组 grid。一个机器人初始位于 左上角（即 grid[0][0]）。机器人尝试移动到 右下角（即 grid[m - 1][n - 1]）。机器人每次只能向下或者向右移动一步。
网格中的障碍物和空位置分别用 1 和 0 来表示。机器人的移动路径中不能包含 任何 有障碍物的方格。
返回机器人能够到达右下角的不同路径数量。
测试用例保证答案小于等于 2 * 109。
示例 1：
输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]] 输出：2 解释：3x3 网格的正中间有一个障碍物。 从左上角到右下角一共有 2 条不同的路径： 1. 向右 -> 向右 -> 向下 -> 向下 2. 向下 -> 向下 -> 向右 -> 向右
示例 2：
输入：obstacleGrid = [[0,1],[0,0]] 输出：1
提示：
-
m == obstacleGrid.length -
n == obstacleGrid[i].length -
1 <= m, n <= 100 -
obstacleGrid[i][j]为0或1
思路分析
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一刷
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二刷
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三刷
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Unique Paths II.
* <p>
* Memory Usage: 40.5 MB, less than 33.84% of Java online submissions for Unique Paths II.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2019-10-26 23:50
*/
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (Objects.isNull(obstacleGrid)
|| obstacleGrid.length == 0
|| obstacleGrid[0].length == 0
|| obstacleGrid[0][0] == 1) {
return 0;
}
obstacleGrid[0][0] = 1;
int m = obstacleGrid.length;
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 0 && obstacleGrid[i - 1][0] == 1) {
obstacleGrid[i][0] = 1;
} else {
obstacleGrid[i][0] = 0;
}
}
int n = obstacleGrid[0].length;
for (int j = 1; j < n; j++) {
if (obstacleGrid[0][j] == 0 && obstacleGrid[0][j - 1] == 1) {
obstacleGrid[0][j] = 1;
} else {
obstacleGrid[0][j] = 0;
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
obstacleGrid[i][j] = 0;
} else {
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
}
}
}
return obstacleGrid[m - 1][n - 1];
}
/**
* Runtime: 1 ms, faster than 23.04% of Java online submissions for Unique Paths II.
* <p>
* Memory Usage: 40.5 MB, less than 35.38% of Java online submissions for Unique Paths II.
*/
public int uniquePathsWithObstaclesMy(int[][] obstacleGrid) {
if (Objects.isNull(obstacleGrid)
|| obstacleGrid.length == 0
|| obstacleGrid[0].length == 0) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
boolean hasObstacle = false;
for (int i = 0; i < n; i++) {
int value = obstacleGrid[0][i];
if (value == 1) {
hasObstacle = true;
obstacleGrid[0][i] = -1;
} else {
if (hasObstacle) {
obstacleGrid[0][i] = 0;
} else {
obstacleGrid[0][i] = 1;
}
}
}
hasObstacle = false;
if (obstacleGrid[0][0] == -1) {
hasObstacle = true;
}
for (int i = 1; i < m; i++) {
int value = obstacleGrid[i][0];
if (hasObstacle) {
obstacleGrid[i][0] = 0;
}
if (value == 1) {
hasObstacle = true;
obstacleGrid[i][0] = -1;
} else {
if (hasObstacle) {
obstacleGrid[i][0] = 0;
} else {
obstacleGrid[i][0] = 1;
}
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
int value = obstacleGrid[i][j];
if (value == 1) {
obstacleGrid[i][j] = -1;
} else {
int top = obstacleGrid[i - 1][j];
if (top == -1) {
top = 0;
}
int left = obstacleGrid[i][j - 1];
if (left == -1) {
left = 0;
}
obstacleGrid[i][j] = top + left;
}
}
}
return Math.max(obstacleGrid[m - 1][n - 1], 0);
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2019-10-26 23:50
*/
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int row = obstacleGrid.length;
int column = obstacleGrid[0].length;
// 1. obstacleGrid 表示走到某个节点一共有多少路径
// 将障碍设置为-1，便于和 0 区分
for (int i = 0; i < row; i++) {
for (int j = 0; j < column; j++) {
if (obstacleGrid[i][j] == 1) {
obstacleGrid[i][j] = -1;
}
}
}
// 3. 第一行也只有一种走法，如果有障碍，则后面的路径为0
for (int i = 0; i < column; i++) {
if (obstacleGrid[0][i] == -1) {
break;
}
obstacleGrid[0][i] = 1;
}
// 3. 第一列只有一种走法，如果有障碍，则后面的路径为0
for (int i = 0; i < row; i++) {
if (obstacleGrid[i][0] == -1) {
break;
}
obstacleGrid[i][0] = 1;
}
printMatrix(obstacleGrid);
// 4. 确定遍历顺序：从左上向右下遍历
for (int i = 1; i < row; i++) {
for (int j = 1; j < column; j++) {
if (obstacleGrid[i][j] == -1) {
continue;
}
// 2. 确定递推公式
int mi = 0;
if (obstacleGrid[i - 1][j] != -1) {
mi = obstacleGrid[i - 1][j];
}
int ni = 0;
if (obstacleGrid[i][j - 1] != -1) {
ni = obstacleGrid[i][j - 1];
}
obstacleGrid[i][j] = mi + ni;
printMatrix(obstacleGrid);
}
}
return obstacleGrid[row - 1][column - 1] == -1 ? 0 : obstacleGrid[row - 1][column - 1];
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-04-18 17:32:04
*/
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
boolean obstacled = false;
boolean firstObstacle = obstacleGrid[0][0] == 1;
for (int i = 0; i < obstacleGrid[0].length; i++) {
if (obstacleGrid[0][i] == 1) {
obstacleGrid[0][i] = 0;
obstacled = true;
} else {
if (obstacled) {
obstacleGrid[0][i] = 0;
} else {
obstacleGrid[0][i] = 1;
}
}
}
obstacled = false;
int rows = obstacleGrid.length;
for (int i = 1; i < rows; i++) {
if (obstacleGrid[i][0] == 1) {
obstacleGrid[i][0] = 0;
obstacled = true;
} else {
if (obstacled || firstObstacle) {
obstacleGrid[i][0] = 0;
} else {
obstacleGrid[i][0] = 1;
}
}
}
for (int i = 1; i < rows; i++) {
for (int j = 1; j < obstacleGrid[i].length; j++) {
if (obstacleGrid[i][j] == 1) {
obstacleGrid[i][j] = 0;
} else {
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
}
}
}
return obstacleGrid[rows - 1][obstacleGrid[rows - 1].length - 1];
}