[LeetCode]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.

开始的时候的答案是这样，会超时的。

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

 你看，这样需要m*n的空间，实际上，我们并不需要保持整个的m*n,只需要保证最外面的一行和一列就好了。

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid .size();
        int n = grid[0].size();
        
        int dp[m],hr[n];
        dp[0] = hr[0] = grid[0][0];
        for(int i = 1; i < m; ++i) 
            dp[i] = dp[i-1] + grid[i][0]; 
        for(int i = 1; i < n; ++i) {
            hr[i] = hr[i-1] + grid[0][i]; 
        }
        // update horizontal line every time for next use
        int tmp;
        for(int i = 1;i<m;++i) {
            hr[0] = dp[i];
            for(int j=1;j<n;++j) {
                hr[j] = min(hr[j-1],hr[j]) + grid[i][j];  
            }
        }
        return hr[n-1];
    }
};

 话说空间对时间有影响么。。。