Minimum Path Sum

Description

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 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]] Output: 7 Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]] Output: 12

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• 0 <= grid[i][j] <= 200

Code

和 Unique Paths ㄧ模一樣。

Time Complexity: $O(nm)$, Space Complexity: $O(nm)$

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, INT_MAX));
        for(int i = 0; i < m; i++) {
            dp[i][0] = grid[i][0];
            if(i > 0)
                dp[i][0] += dp[i - 1][0];
        }
            
        for(int i = 0; i < n; i++) {
            dp[0][i] = grid[0][i];
            if(i > 0)
                dp[0][i] += dp[0][i - 1];
        }
            
        
        for(int i = 1; i < m; i++) {
            for(int j = 1; j < n; j++) {
                dp[i][j] = min(dp[i][j], dp[i][j - 1] + grid[i][j]);
                dp[i][j] = min(dp[i][j], dp[i - 1][j] + grid[i][j]);
            }
        }
 
        return dp[m - 1][n - 1];
    }
 
};
 

Source

• Minimum Path Sum - LeetCode