力扣 题目63- 不同路径II

题目

题解

基本思想在62题 我们只要让障碍物在的格子取0即可

代码

#include<iostream>
#include<vector>
using namespace std;

class Solution {
    public:
        int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
            int m = obstacleGrid.size();
            int n = obstacleGrid[0].size();
            vector<vector<int>> dp(m, vector<int>(n, 0));
            //一列 所以必定只有一条 路线
            for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
            //一行 所以必定只有一条 路线
            for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;
            for (int i = 1; i < m; i++) {
                for (int j = 1; j < n; j++) {
                    if (obstacleGrid[i][j] == 1) continue;
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
            return dp[m - 1][n - 1];
        }
    };

int main() {
    Solution sol;
    vector<vector<int>> obstacleGrid = {{0,0,0} ,{0,1,0},{0,0,0},{0,0,0} };
    int result = sol.uniquePathsWithObstacles(obstacleGrid);
    cout << result << endl;
}