LeetCode 63. Unique Paths II

最新推荐文章于 2025-09-10 15:24:29 发布

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原文链接：https://segmentfault.com/a/1190000012791207

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#数据结构与算法

解决在网格中寻找从起点到终点的唯一路径数量问题，当网格中存在障碍物时，使用动态规划方法进行求解。

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.

大致题意

在 unique paths I 基础上增加障碍物

题解

题目类型：动态规划

转移方程为：

path[i][j] = obstacleGrid[i][j] ? 0 : path[i - 1][j] + path[i][j - 1];

解如下

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