LeetCode //C – 64. Minimum Path Sum

64. 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.
 

Example 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• 0 <= grid[i][j] <= 200

From: LeetCode
Link: 64. Minimum Path Sum

Solution:

Ideas：

This function assumes that the memory for grid has already been allocated and that gridSize and gridColSize are correctly set to reflect the dimensions of grid. The min function is a helper to find the minimum of two numbers.

The function calculates the minimum path sum in a bottom-up manner, filling in the dp table from the top–left to the bottom-right. After calculating the minimum path sum, it cleans up the allocated memory for dp and returns the result.

Code:

int minPathSum(int** grid, int gridSize, int* gridColSize) {
    // The gridSize is the number of rows, and gridColSize[0] is the number of columns.
    int i, j;
    
    // Allocate space for the dp matrix
    int **dp = (int **)malloc(gridSize * sizeof(int *));
    for(i = 0; i < gridSize; i++) {
        dp[i] = (int *)malloc(gridColSize[0] * sizeof(int));
    }
    
    // Initialize the top-left corner
    dp[0][0] = grid[0][0];
    
    // Fill the first row (only right moves are possible)
    for(j = 1; j < gridColSize[0]; j++) {
        dp[0][j] = dp[0][j - 1] + grid[0][j];
    }
    
    // Fill the first column (only down moves are possible)
    for(i = 1; i < gridSize; i++) {
        dp[i][0] = dp[i - 1][0] + grid[i][0];
    }
    
    // Fill the rest of the dp matrix
    for(i = 1; i < gridSize; i++) {
        for(j = 1; j < gridColSize[0]; j++) {
            dp[i][j] = grid[i][j] + min(dp[i - 1][j], dp[i][j - 1]);
        }
    }
    
    // The bottom-right corner has the result
    int result = dp[gridSize - 1][gridColSize[0] - 1];
    
    // Free the dp matrix
    for(i = 0; i < gridSize; i++) {
        free(dp[i]);
    }
    free(dp);
    
    return result;
}

// Helper function to find the minimum of two numbers
int min(int a, int b) {
    return (a < b) ? a : b;
}

代码007普通

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