LeetCode 64. Minimum Path Sum

[woodyiiiiiii]

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:

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

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解法1:

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

一维数组:

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

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参考资料:
LeetCode原题