64. Minimum Path Sum

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.

Example 1:

[[1,3,1],
 [1,5,1],
 [4,2,1]]

Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum. 

解析

// add 64. Minimum Path Sum
class Solution_64 {
public:
    int minPathSum(vector<vector<int>>& grid) {

        int m = grid.size();
        int n = grid[0].size();

        vector<vector<int>> dp(m, vector<int>(n, 0));
        
        for (int i = 0; i < m;i++)
        {
            for (int j = 0; j < n;j++)
            {
                if (i==0&&j==0)
                {
                    dp[i][j] = grid[0][0];
                }else if (i==0)
                {
                    dp[i][j] = dp[i][j - 1]+grid[i][j];
                }
                else if (j==0)
                {
                    dp[i][j] = dp[i-1][j] + grid[i][j];
                }
                else
                {
                    dp[i][j] = min(dp[i-1][j], dp[i][j - 1]) + grid[i][j];
                }
                
            }
        }
        return dp[m-1][n-1];
    }
};

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

题目来源

• 64. Minimum Path Sum