算法分析与设计——LeetCode:63. Unique Paths II

题目

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.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    }
};

思路

这道题跟上一道Unique Paths I类似，不同在于需考虑当前位置是否为障碍物，如果为障碍物，就将当前位置的路径数量直接赋值为0。

代码

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        int grid[100][100];
        grid[0][0] = 1;
        for (int i = 0; i < m; i++) {
            grid[i][0] = 0;
        }
        for (int i = 0; i < n; i++) {
            grid[0][i] = 0;
        }
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                break;
            }
            grid[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            grid[0][i] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if(obstacleGrid[i][j] == 1) {
                    grid[i][j] = 0;
                } else {
                    grid[i][j] = grid[i-1][j]+grid[i][j-1];
                }
            }
        }
        return grid[m-1][n-1];
    }
};