算法刷题记录 Day34

Date: 2024.03.30

lc 63. 不同路径II

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& G) {// dp[i][j]表示从（0，0）到（i,j）的路径数；// dp[i][j] = dp[i-1][j] + dp[i][j-1] if G(i,j)!=1 else dp[i][j] = 0;// dp[0][j] = dp[0][j-1]; ifG(0,j) = 1,dp[0][j] = 0;int m = G.size();int n = G[0].size();vector<vector<int>> dp(m, vector<int>(n, 0));for(int i=0; i<m; i++){if(G[i][0] == 1)break;dp[i][0] = 1;}for(int j=0; j<n; j++){if(G[0][j] == 1)break;dp[0][j] = 1;}for(int i=1; i<m; i++){for(int j=1; j<n; j++){if(G[i][j] == 1){dp[i][j] = 0;}else{dp[i][j] = dp[i-1][j] + dp[i][j-1];//cout<<"i:"<<i<<" j:"<<j<<" dp:"<<dp[i][j]<<endl;}}}return dp[m-1][n-1];}
};

lc 62. 不同路径

class Solution {
public:int uniquePaths(int m, int n) {// dp[i][j] 表示从（0，0）到（i,j）的路径数；// dp[i][j] = dp[i-1][j] + dp[i][j-1];// dp[i][0] = 1; dp[0][j] = 1;vector<vector<int>> dp(m, vector<int>(n, 1));for(int i=1; i<m; i++){for(int j=1; j<n; j++){dp[i][j] = dp[i-1][j] + dp[i][j-1];//cout<<"(i, j):"<<i<<","<<j<<"dp[i][j]:"<<dp[i][j]<<endl;}}return dp[m-1][n-1];}
};