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:

Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7

Explanation: Because the path 1→3→1→1→1 minimizes the sum.

DP： 答案可以存在二维数组中

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

优化： 可以使用一维数组代替二维数组 因为用到的前面的答案只是它的相邻两格的答案，所以可以用一个一维数组进行覆盖存储。

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