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  • LeetCode Unique Paths II

    class Solution {
    public:
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
            vector<vector<int> >& dp = obstacleGrid;
            if (dp.empty() || dp[0].empty()) return 0;
            dp[0][0] = (dp[0][0] == 1) ? 0 : 1;
            for (int i=1; i<dp.size(); i++) {
                dp[i][0] = (dp[i][0] == 1) ? 0 : dp[i-1][0];
            }
            for (int i=1; i<dp[0].size(); i++) {
                dp[0][i] = (dp[0][i] == 1) ? 0 : dp[0][i-1];
            }
            for (int i=1; i<dp.size(); i++) {
                for (int j=1; j<dp[i].size(); j++) {
                    dp[i][j] = (dp[i][j] == 1) ? 0 : dp[i-1][j] + dp[i][j-1];
                }
            }
            return dp.back().back();
        }
    };

    和Unique Paths一样只不过这回不太能用公式直接得出了,在动态规划时依据提供的障碍信息确定如何选取子问题的解,方便起见直接将入参用作dp数组

    第二轮:

    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.

    如果当前位置有障碍那么该处的dp数组元素直接为0,其余和unique path一样

    class Solution {
    public:
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
            int m = obstacleGrid.size();
            if (m < 1) return 0;
            int n = obstacleGrid[0].size();
            if (n < 1) return 0;
            
            int* dp = new int[n+1];
            for (int i=0; i<=n; i++) dp[i] = 0;
            if (obstacleGrid[0][0] != 1) dp[1] = 1;
            
            for (int i=0; i<m; i++) {
                for (int j=1; j<=n; j++) {
                    if (obstacleGrid[i][j-1] != 1) {
                        dp[j] = dp[j] + dp[j-1];
                    } else {
                        dp[j] = 0;
                    }
                }
            }
            return dp[n];
        }
    };
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  • 原文地址:https://www.cnblogs.com/lailailai/p/3601210.html
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