64. Minimum Path Sum 动态规划

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:

Example:

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.

answer:

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

relative point get√：

hint :

动态规划，边界情况，更新条件