【Minimum Path Sum】cpp

题目：

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.

代码：

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

tips：

典型的“DP+滚动数组”，时间复杂度O(m*n)，空间复杂度O(n)。

=============================================

第二次，用偷懒的做法了，二维dp直接写了。

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
            if ( grid.empty() ) return ;
            int dp[grid.size()][grid[].size()];
            fill_n(&dp[][], grid.size()*grid[].size(), );
            dp[][] = grid[][];
            for ( int i=; i<grid[].size(); ++i ) dp[][i] = dp[][i-]+grid[][i];
            for ( int i=; i<grid.size(); ++i ) dp[i][] = dp[i-][]+grid[i][];
            for ( int i=; i<grid.size(); ++i )
            {
                for ( int j=; j<grid[i].size(); ++j )
                {
                    dp[i][j] = min(dp[i][j-],dp[i-][j])+grid[i][j];
                }
            }
            return dp[grid.size()-][grid[].size()-];
    }
};