leetcode64.最小路径和（中等）

解法：动态规划
状态转移方程为：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 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 < n; ++i) {
            dp[0][i] = dp[0][i - 1] + grid[0][i];
        }
        for (int i = 1; i < m; ++i) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }
        return dp[m - 1][n - 1];
    }
};

标签：leetcode64,int,路径,中等,++,vector,grid,dp,size
来源： https://blog.csdn.net/zhangjiaji111/article/details/120967884