【leetcode刷题】T157-不同路径 II

示例 1:
输入:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
输出: 2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        # 特殊情况
        if not obstacleGrid or not obstacleGrid[0]:
            return 0

        # dp[i+1][j+1] = dp[i][j+1] + dp[i-1][j]
        dp = [[0] * len(obstacleGrid[0]) for i in range(len(obstacleGrid))]
        dp[0][0] = 1
        for i in range(len(obstacleGrid)):
            for j in range(len(obstacleGrid[0])):
                if obstacleGrid[i][j] == 1:
                    dp[i][j] = 0
                else:
                    if i > 0:
                        dp[i][j] += dp[i-1][j]
                    if j > 0:
                        dp[i][j] += dp[i][j-1]
        return dp[-1][-1]

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0)
            return 0;

        // dp[i][j] = dp[i-1][j] + dp[i][j-1]
        // ==> dp[j] += dp[j-1]
        vector<double> dp(obstacleGrid[0].size(), 0);
        dp[0] = 1;
        for(int i=0; i < obstacleGrid.size(); i++){
            for(int j=0; j < obstacleGrid[0].size(); j++){
                if(obstacleGrid[i][j] == 1)
                    dp[j] = 0;
                else{
                    if(j > 0)
                        dp[j] += dp[j-1];
                }
            }
        }
        return dp.back();
    }
};