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Diffstat (limited to 'unique_paths2.cpp')
-rw-r--r-- | unique_paths2.cpp | 39 |
1 files changed, 0 insertions, 39 deletions
diff --git a/unique_paths2.cpp b/unique_paths2.cpp deleted file mode 100644 index fedb826..0000000 --- a/unique_paths2.cpp +++ /dev/null @@ -1,39 +0,0 @@ -// 63. Unique Paths II -#include "leetcode.h" - -/* - You are given an m x n integer array grid. There is a robot initially located - at the top-left corner (i.e., grid[0][0]). The robot tries to move to the - bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either - down or right at any point in time. - - An obstacle and space are marked as 1 or 0 respectively in grid. A path that - the robot takes cannot include any square that is an obstacle. - - Return the number of possible unique paths that the robot can take to reach - the bottom-right corner. - - The testcases are generated so that the answer will be less than or equal to 2 - * 109. - */ - -class Solution { -public: - int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) { - int m = obstacleGrid.size(), n = obstacleGrid[0].size(); - vvd dp(m + 1, vector<int>(n + 1, 0)); - dp[0][1] = 1; - for (int i = 1; i <= m; ++i) - for (int j = 1; j <= n; ++j) - if (!obstacleGrid[i - 1][j - 1]) - dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; - return dp[m][n]; - } -}; - -int main() { - Solution obj; - vvd og1 = {{0, 0, 0}, {0, 1, 0}, {0, 0, 0}}, og2 = {{0, 1}, {0, 0}}; - printf("%d\n", obj.uniquePathsWithObstacles(og1)); // expect: 2 - printf("%d\n", obj.uniquePathsWithObstacles(og2)); // expect: 1 -} |