[leetcode] Unique Paths II

最新推荐文章于 2025-01-13 17:23:47 发布

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本文探讨了在存在障碍物的情况下计算从左上角到右下角的不同路径数量的问题。提供了两种解决方案，一种使用二维动态规划，另一种则通过优化内存消耗采用一维数组实现。这两种方法都有效地解决了问题，并在LeetCode的“Unique Paths II”题目中得到了验证。

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From : https://leetcode.com/problems/unique-paths-ii/

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.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m=obstacleGrid.size(), n=obstacleGrid[0].size();
        vector<vector<int>> steps(m, vector<int>(n, 0));
		int flag = 1;
        for(int i=0; i<m; i++) {
			flag = flag&(!obstacleGrid[i][0]);
			steps[i][0] = flag;
		}
		flag = 1;
        for(int j=0; j<n; j++) {			
			flag = flag&(!obstacleGrid[0][j]);
			steps[0][j] = flag;
		}
        for(int i=1; i<m; i++) {
            for(int j=1; j<n; j++) {
                steps[i][j] = (!obstacleGrid[i][j])*(steps[i-1][j]+steps[i][j-1]);
            }
        }
        return steps[m-1][n-1];
    }
};

优化：

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

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0) {
            return 0;
        }
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[] v = new int[n];
        for(int i=0; i<n && obstacleGrid[0][i]==0; ++i) {
            v[i] = 1;
        }
        
        int flag = 1-obstacleGrid[0][0];
        for(int i=1; i<m; ++i) {
            if(flag==0 || obstacleGrid[i][0]==1) {
                flag = 0;
            }
            v[0] = flag;
            for(int j=1; j<n; ++j) {
                v[j] = (1-obstacleGrid[i][j])*(v[j]+v[j-1]);
            }
        }
        
        return v[n-1];
    }
}