Problem:

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.

Idea:

DP

Solution 1:

C++ / Recursion with memoization

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua
// Runtime: 9 ms
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        if (m == 0) return 0;
        int n = grid[0].size();
        
        s_ = vector<vector<int>>(m, vector<int>(n, 0));
        
        return minPathSum(grid, n - 1, m - 1, n, m);
    }    
private:
    long minPathSum(const vector<vector<int>>& grid, 
                   int x, int y, int n, int m) {        
        if (x == 0 && y == 0) return grid[y][x];
        if (x < 0 || y < 0) return INT_MAX;
        if (s_[y][x] > 0) return s_[y][x];
        
        int ans = grid[y][x] + min(minPathSum(grid, x - 1, y, n, m),
                                   minPathSum(grid, x, y - 1, n, m));
        return s_[y][x] = ans;
    }
    
    vector<vector<int>> s_;
};

Solution 2:

C++ / DP

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua
// Runtime: 9 ms
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        if (m == 0) return 0;
        int n = grid[0].size();
        
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j) {
                if (i == 0 && j == 0) continue;
                if (i == 0) 
                    grid[i][j] += grid[i][j - 1];
                else if (j == 0)
                    grid[i][j] += grid[i - 1][j];
                else
                    grid[i][j] += min(grid[i][j - 1], grid[i - 1][j]);
            }
        
        return grid[m - 1][n - 1];
    }    
};

Related Problems:

• [解题报告] LeetCode 113. Path Sum II

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