leetcode 063 Unique Paths II

最新推荐文章于 2024-07-26 10:47:39 发布

walter1990

最新推荐文章于 2024-07-26 10:47:39 发布

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本文探讨了在网格中寻找从起点到终点的独特路径数量问题，并在此基础上增加了障碍物元素。通过动态规划的方法，解决如何计算存在障碍物时的独特路径数量。

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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.

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

		for(int i=1; i < m; i++) {
		    for(int j=1; j < n; j++) {
		        
		        if(obstacleGrid[i][j]) matrix[i][j] = 0;
		        else
		            matrix[i][j] = matrix[i][j-1]+matrix[i-1][j];
		    }
		}
		
		return matrix[m-1][n-1];
    }
};