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1. Topic analysis

Topic Link: 64. Minimum Path Sum - Leetcode

This question is not difficult to understand, starting from the upper right corner,

You can only go to the right or down, and calculate the minimum path sum to the lower right corner.

2. Algorithm principle

1. Status representation

dp[ i ][ j ] means the minimum path sum from the starting point to the position [ i, j ].

2. State transition equation

Divide the problem according to the latest step, and there are two cases to the [i, j] position,

One is from above: dp[ i - 1 ][ j ] + g[ i ][ j ]

One is from the left: dp[ i ][ j - 1 ] + g[ i ][ j ]

And what we require is the minimum path sum, so the state transition equation is:

dp[ i ][ j ] = min( dp[ i - 1 ][ j ]，dp[ i ][ j - 1 ] ) + g[ i ][ j ]

3. Initialization

The initialization is to prevent out of bounds, so we only need to add the upper line and the left line,

However, the value filled in must ensure that it does not affect the subsequent form filling, so we can initialize the entire table to positive infinity,

Then, in order to ensure that there is no problem with filling out the form for the first time, just initialize the position above the first position to 0.

4. Filling order

top to bottom, left to right

5. Return value

Return the minimum path sum in the lower right corner.

3. Code writing

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

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[Learning Dynamic Programming] Minimal Path Sum (9)