[LeetCode]63. Unique Paths II

题目描述

是 [LeetCode]62. Unique Paths的扩展
增加了路障

思路

动态规划
过程中需要考虑该点是否可以通过，对于不可通过的点，在dp数组中直接置为0；
在初始状态时也要考虑，若终点不可到达，直接置为0，若在与终点处于同一直线的点不可到达，需要将这一点前的所有点置为0
例如
当前状态为

ob 0 1 2 0 0 0 0 1 0 1 0

则对应的dp数组为

dp 0 1 2 0 1 1 1 1 0 0 1

代码

#include <iostream>#include <vector>using namespace std;class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int row = obstacleGrid.size(),            col = obstacleGrid[0].size();        vector<vector<int>> dp(row, vector<int>(col));        for(int i = row - 1; i >= 0; --i)            for (int j = col - 1; j >= 0; --j) {                if (i == row - 1 && j == col - 1)                    dp[i][j] = obstacleGrid[i][j] == 0 ? 1 : 0;                else if (i == row - 1)                    dp[i][j] = dp[i][j + 1] && (!obstacleGrid[i][j]);                else if (j == col - 1)                    dp[i][j] = dp[i + 1][j] && (!obstacleGrid[i][j]);                else                    dp[i][j] = obstacleGrid[i][j] == 0 ? dp[i][j + 1] + dp[i + 1][j] : 0;            }        return dp[0][0];    }};int main() {    vector<vector<int>> obs = { {0,0,0},{0,1,0},{0,0,0} };    Solution s;    cout << s.uniquePathsWithObstacles(obs) << endl;    system("pause");    return 0;}