【Leetcode】63. Unique Paths II（动态规划）

最新推荐文章于 2024-10-01 18:52:56 发布

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最新推荐文章于 2024-10-01 18:52:56 发布

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分类专栏：【LeetCode】刷题记录文章标签：动态规划

本文链接：https://blog.youkuaiyun.com/qq_29600137/article/details/88843379

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本文介绍了一种算法，用于计算在一个存在障碍物的网格中，从起点到终点的不同路径数量。该算法采用动态规划方法，考虑了只能向下或向右移动的限制条件。

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

题目大意：

给出一个二维矩阵，其中包含0和1两种状态，0代表可以通过，1则代表无法通过；输出从左上角到右下角的路径个数，注意此时只可以向右或者向下走。

解题思路：

题目与62题类似，只是多出了一个无法通过的状态。按照62题的思路，我们希望计算出到达每个位置的个数，然后动态规划。对本题而言，我们可以看出在标注为1的位置是永远无法到达的，所以在遍历时应该将此处的个数设为0。

同时在初始化第一行第一列时，我们需要注意，因为机器人只允许往下走不允许往上，所以我们在初始化第一行第一列时应该注意，如果出现1则本位置和之后的所有位置都无法到达，即标注其值为0.

注：此题使用c++会超出int的范围，将变量设为 long long int。

62题题解：https://blog.youkuaiyun.com/qq_29600137/article/details/88734291

附C++ faster than 100.00% 代码

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        
        if(obstacleGrid[0][0]==1)
            return 0;
        
        long long int n,m;
        n = obstacleGrid.size();
        m = obstacleGrid[0].size();
        long long int map[n][m];
        bool state_row = false;
        for(int i=0;i<m;i++){
            if (obstacleGrid[0][i] == 1 || state_row){
                map[0][i] = 0;
                state_row = true;
                continue;
            }
			map[0][i] = 1;
		} 
        
        bool state_line = false;
		for(int i=0;i<n;i++){
            if (obstacleGrid[i][0] == 1 || state_line){
                map[i][0] = 0;
                state_line = true;
                continue;
            }
			map[i][0] = 1;
		}
        
        for(int i=1;i<n;i++){
			for(int j=1;j<m;j++){
                
                if(obstacleGrid[i][j] == 1){
                    map[i][j] = 0;
                    continue;
                }
                
                if(obstacleGrid[i][j-1]!=1 && obstacleGrid[i-1][j]!=1)
				    map[i][j] = map[i][j-1] + map[i-1][j];
                else{
                    if(obstacleGrid[i][j-1]==1 && obstacleGrid[i-1][j]!=1)
                        map[i][j] = map[i-1][j];
                    else if(obstacleGrid[i][j-1]!=1 && obstacleGrid[i-1][j]==1){
                        map[i][j] = map[i][j-1];
                    }else{
                        map[i][j] = 0;
                    }
                }
			}
		}
        return map[n-1][m-1];
    }
};