63. Unique Paths II

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探讨如何使用动态规划解决二维网格中避开障碍物寻找从起点到终点的不同路径数量问题。通过初始化边界条件并递推填充状态矩阵来计算可能路径总数。

动态规划的问题，属于坐标型动态规划（二维）

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.

java

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