Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
1 class Solution { 2 public: 3 int caluniquePaths(vector<vector<int>>&grid,vector<vector<int>>&mark,int m, int n) { 4 int down = -1, right = -1; 5 if (mark[m][n] != -1)return mark[m][n]; 6 int orim = grid.size(), orin = grid[0].size(); 7 if (m > 1) 8 down = caluniquePaths(grid,mark, m - 1, n); 9 if (n > 1) 10 right = caluniquePaths(grid,mark, m, n - 1); 11 if (down == -1)down = right; 12 if (right == -1)right = down; 13 mark[m][n] = grid[orim-m][orin-n]+min(down,right); 14 return mark[m][n]; 15 } 16 int minPathSum(vector<vector<int>>& grid) { 17 int m = grid.size(), n = grid[0].size(); 18 vector<vector<int>>mark(m+1, vector<int>(n+1, -1)); 19 mark[1][1] = grid[m-1][n-1]; 20 return caluniquePaths(grid,mark, m, n); 21 } 22 };
用的改的上一题的题解，但偏慢，不如改成dp快，估计是调用比较慢
1 class Solution { 2 public: 3 int minPathSum(vector<vector<int>>& grid) { 4 int m = grid.size(), n = grid[0].size(); 5 vector<vector<int>>dp(m, vector<int>(n, INT_MAX)); 6 dp[0][0] = grid[0][0]; 7 for(int i=0;i<m;i++) 8 for (int j = 0; j < n; j++) { 9 if (i == 0 && j == 0)continue; 10 if (i == 0) { 11 dp[i][j] = min(dp[i][j - 1] + grid[i][j], dp[i][j]); 12 } 13 else if (j == 0) { 14 dp[i][j] = min(dp[i - 1][j] + grid[i][j], dp[i][j]); 15 } 16 else { 17 dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j], dp[i][j]); 18 } 19 } 20 return dp[m - 1][n - 1]; 21 } 22 };