算法分析與設計——LeetCode Problem.63 Unique Paths II

題目鏈接

問題描述

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.

解題思路

這題與Leetcode 62解決方法相似，可以新建一個與grid相同size的矩陣path。需要注意的是grid的最左邊一列grid[i][0]和最上邊一排grid[0][i]：從grid的左上角grid[0][0]開始，分別沿最左邊和最上邊遍歷，直到遍歷完整一行/列或者遇到數字1，在此之前要把path相應位置賦值爲1。之後根據遞推式path[i][j] = path[i-1][j] + path[i][j-1]遍歷path矩陣即可，若所遍歷的位置path[i][j]在grid矩陣中的相應位置grid[i][j]的值爲1，則將path[i][j]的值賦爲0即可。

代碼如下

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