LeetCode进阶之路（Unique Paths II）

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本文介绍了一种求解在网格中存在障碍物时从起点到终点的不同路径数量的算法。该算法首先将障碍物标记为特殊值，然后通过动态规划的方法计算每个位置到达终点的路径数。最终返回右下角的值即为不同路径的数量。

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.

题目：这道题是上一题的升级版，中间加了一个障碍，其实只要把障碍地方的解变为 0 就好。

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int row = obstacleGrid.length;
        if(row < 1 ) return 0;
        int col = obstacleGrid[0].length;
        if(col < 1) return 0;
        
        if(obstacleGrid[0][0] == 1|| obstacleGrid[row-1][col-1]==1) return 0;
        
        for(int i = 0;i < row;i++) {
            for(int j = 0;j < col;j++) {
                if(obstacleGrid[i][j] == 1){
                    obstacleGrid[i][j] = -1;
                }
            }
        }
        
        for(int i = 0;i < row;i++){
            if(obstacleGrid[i][0] == -1) {
                break;
            } else {
                obstacleGrid[i][0] = 1;
            }
        }
        
        for(int i = 0;i < col;i++){
            if(obstacleGrid[0][i] == -1) {
                break;
            } else {
                obstacleGrid[0][i] = 1;
            }
        }
        
        for(int i = 1;i < row;i++) {
            for(int j = 1;j < col;j++) {
                if(obstacleGrid[i][j] != -1) {
                    if(obstacleGrid[i-1][j] ==-1 && obstacleGrid[i][j-1] == -1) {
                        continue;
                    }else if(obstacleGrid[i-1][j] ==-1) {
                        obstacleGrid[i][j] = obstacleGrid[i][j-1];
                    } else if(obstacleGrid[i][j-1] == -1) {
                        obstacleGrid[i][j] = obstacleGrid[i-1][j];
                    } else {
                        obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
                    }
                }
            }
        }
        return obstacleGrid[row-1][col-1];
    }