64.Minimum Path Sum

Posted 2020-12-07 luo-c

给定一个矩阵，矩阵中的元素表示成本，求，从矩阵左上角到右下角最小的成本路线，每一次只能向右或者向下走。

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：
此题跟62题类似，不同点在于，此题计算最小成本，则将DP中存储每一个节点的最小成本即可。对于DP的初始，两边的值需要单独初始化，其余的点：dp[i][j] = min(dp[i-1][j] , dp[i][j-1]) + grid[i][j].

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