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*62

leetcode quaestio electronica

Utriusque cellae valor supra et ad sinistram determinatur, ideo supremae et sinistrae cellae primo initialized oportet. Quia quaestio postulat, ut solum deorsum et dextrorsum movearis, 1 cellae supremae et sinistrae initialized ad 1 .

temporis complexionem; $O****(m\ast n)$
Spatium complexionis: $O****(m\ast n)$

// c++
class Solution {
public:
    
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(m, vector(n,0));
        for(int i=0; i<m; i++) dp[i][0] = 1;
        for(int j=0; j<n; j++) dp[0][j] = 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];
            }
        }
        return dp[m-1][n-1];
    }
};
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63. diuersae semitae II

leetcode quaestio electronica

Similis est idea praecedenti quaestioni, et adduntur impedimenta ex praecedenti quaestione. In dp ordinata numerus semitarum ad cellulam (i, j) significat, et obstaculum positio 0 in dp ordinata est. Incipiens ab initializatione primi ordinis et columnae, si impedimentum in viis apparet, cellae post impedimentum non attingi possunt, et valores ordinatae dp respondentes omnes 0 erunt.

temporis complexionem; $O****(m\ast n)$
Spatium complexionis: $O****(m\ast n)$

// c++
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        vector<vector<int>> dp(obstacleGrid.size(), vector(obstacleGrid[0].size(), 0));
        // 标志单元格是否可到达
        bool flag = true;
        // 初始化第一行
        for(int i=0; i<obstacleGrid.size(); i++){
            if(obstacleGrid[i][0] == 1) {
            	// 可以直接break
            	flag = false;
            }
            if(flag) dp[i][0] = 1;
            else dp[i][0] = 0;
        }
        // 标志单元格是否可到达
        flag = true;
        // 初始化第一列
        for(int j=0; j<obstacleGrid[0].size(); j++){
            if(obstacleGrid[0][j] == 1) {
            	// 可以直接break
            	flag = false;
            }
            if(flag) dp[0][j] = 1;
            else dp[0][j] = 0;
        }
        // 计算dp数组
        for(int i=1; i<obstacleGrid.size(); i++){
            for(int j=1; j<obstacleGrid[0].size(); j++){
                if(obstacleGrid[i][j]==1) continue;
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[obstacleGrid.size()-1][obstacleGrid[0].size()-1];
    }
};
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