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.

Tip: The idea is similar to unique path, calculate left and up, and then calculate others. This process can be simplified from two columns to one column, saving space.

Answer:

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