【LeetCode】64. Minimum path sum

topic

Given a   grid  of non-negative integers  , find a path from the upper left corner to the lower right corner that minimizes the sum of numbers along the path.m x ngrid

Note: Only move down or right one step at a time.

Example 1:

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

Example 2:

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

hint:

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

answer

source code

class Solution {
    public int minPathSum(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] dp = new int[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; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j];
            }
        }

        return dp[m - 1][n - 1];
    }
}

Summarize

Another dynamic programming, this time I wrote it myself! Use grid[i][j] to represent the minimum path sum from (0,0) to (m,n), equal to the minimum value in dp[i][j - 1] and dp[i - 1][j] plus grid[i][j].