【063-Unique Paths II（唯一路径问题II）】
【LeetCode-面试算法经典-Java实现】【全部题目文件夹索引】
原题
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.
题目大意
唯一路径问题兴许。假设路径中有障碍，求总的路径的种数。
【唯一路径问题】
注意：网格的行数和列数都不超过100
解题思路
採用分治求解方法
用一个ｍ*n的组数A保存结果。
对于A数组中的元素有。
1、当x=0或者y=0时。而且(x, y)位置无障碍。
有A[x][y] = 1。 有障碍就是0
2、当x>=1而且y>=1时，而且(x, y)位置无障碍。
有A[x][y] = A[x-1][y]+A[x][y-1]。 有障碍就是0
3、所求的结点就是A[m-1][n-1]。
代码实现
算法实现类
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
// 输入校验
if (obstacleGrid == null || obstacleGrid.length < 1 || obstacleGrid[0].length < 1
|| obstacleGrid[0][0] == 1
|| obstacleGrid[obstacleGrid.length - 1][obstacleGrid[0].length - 1] == 1) {
return 0;
}
int rows = obstacleGrid.length;
int cols = obstacleGrid[0].length;
int[][] result = new int[rows][cols];
// 第一个位置有多少种方法。无障碍就是1种，有障碍就是0种
result[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0;
for (int i = 1; i < cols; i++) {
result[0][i] = obstacleGrid[0][i] == 0 ? result[0][i - 1] : 0;
}
for (int i = 1; i < rows; i++) {
result[i][0] = obstacleGrid[i][0] == 0 ? result[i - 1][0] : 0;
}
for (int i = 1; i < rows; i++) {
for (int j = 1; j < cols; j++) {
result[i][j] = obstacleGrid[i][j] == 0 ?
result[i - 1][j] + result[i][j - 1] : 0;
}
}
return result[rows - 1][cols - 1];
}
// 使用递归方法会超时
public int uniquePathsWithObstacles2(int[][] obstacleGrid) {
// 输入校验
if (obstacleGrid == null || obstacleGrid.length < 1 || obstacleGrid[0].length < 1
|| obstacleGrid[obstacleGrid.length - 1][obstacleGrid[0].length - 1] == 1) {
return 0;
}
int[] result = {0};
solve(obstacleGrid, 0, 0, result);
return result[0];
}
public void solve(int[][] grid, int row, int col, int[] sum) {
// 到达终点
if (row == grid.length - 1 && col == grid[0].length - 1) {
sum[0]++;
}
// 没有到终点，点在棋盘内。而且当前位置不是
else if (row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && grid[row][col] == 0) {
// 往右走
solve(grid, row, col + 1, sum);
// 往下走
solve(grid, row + 1, col, sum);
}
}
}
评測结果
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