[LeetCode] 63. Unique Paths II

[LeetCode] 63. Unique Paths II

题目描述

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.

分析

这个问题和Unique Path类似，但是要注意几个问题：
- 在对第一列和第一行进行初始化时，当遇到障碍物时，从障碍物往后的格子dp的值都为0。
- 在对其他列进行计算时，遇到障碍物则该格子不可以进行计算。
直接计算组合数：

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