【leetcode】Unique Paths II

最新推荐文章于 2019-05-17 12:08:12 发布

原创最新推荐文章于 2019-05-17 12:08:12 发布 · 288 阅读

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#c #leetcode #算法 #动态规划

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本文探讨了一个3x3网格中存在障碍物的情况下，从左上角到右下角的不同路径数量计算问题。通过动态规划的方法，文章提供了一种高效的解决方案，并给出具体实现代码。

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.

Accept: 2ms, 动态规划

int dp[100][100];

int uniquePathsWithObstacles(int **obstacleGrid, int m, int n){
  if (obstacleGrid[m-1][n-1] == 1 || obstacleGrid[0][0] == 1) {
    return 0;
  }
  for (int i = m - 1; i >= 0; --i) {
    for (int j = n - 1; j >= 0; --j) {
      if (i == m - 1 && j == n - 1) {
        dp[i][j] = 1;
        continue;
      }
      dp[i][j] = 0;
      if (i < m - 1 && obstacleGrid[i + 1][j] == 0) {
        dp[i][j] += dp[i + 1][j];
      }
      if (j < n - 1 && obstacleGrid[i][j + 1] == 0) {
        dp[i][j] += dp[i][j + 1];
      }
    }
  }
  return dp[0][0];
}