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.

思路

动态规划问题

c++11

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

这里 decltype(grid.size()) 和 decltype(grid[0].size()) 的类型为 size_t，也可直接使用。

动态规划

参见知乎上的讨论。