64. Minimum Path Sum

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.

my code:

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

Runtime: 12 ms, faster than 20.71% of C++ online submissions for Minimum Path Sum.

 

永远渴望，大智若愚（stay hungry, stay foolish）