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题目链接:LeetCode 62.不同路径
思路:
二维数组dp,简单
class Solution { public: int uniquePaths(int m, int n) { if( m==1 || n==1 ) return 1; vector<vector<int>> dp(m,vector<int>(n,0)); dp[0][0] = 1; for(int i = 1; i<m; i++ ){ for(int j = 1; j<n; j++ ){ if(i == 1){ dp[i-1][j] = 1; } if(j == 1){ dp[i][j-1] = 1; } dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; } };
题目链接:LeetCode 63. 不同路径 II
思路:
分4种情况,写法需要优化
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)); if (obstacleGrid[0][0]) return 0; dp[0][0] = 1; for(int i=1; i<m; i++){ if (obstacleGrid[i][0] == 1){ for(; i<m; i++) dp[i][0] = 0; } else dp[i][0] = 1; } for(int j=1; j<n; j++){ if (obstacleGrid[0][j] == 1){ for(; j<n; j++) dp[0][j] = 0; } else dp[0][j] = 1; } for(int i = 1; i<m; i++){ for(int j = 1; j<n; j++){ if (obstacleGrid[i-1][j] == 1 && obstacleGrid[i][j-1] != 1) dp[i][j] = dp[i][j-1]; else if (obstacleGrid[i][j-1] == 1 && obstacleGrid[i-1][j] != 1) dp[i][j] = dp[i-1][j]; else if ((obstacleGrid[i-1][j] == 1 && obstacleGrid[i][j-1] == 1) || (obstacleGrid[i][j]==1)) dp[i][j] = 0; else dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; } };
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