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1 | 1 | package com.fishercoder.solutions;
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2 | 2 |
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3 |
| -/**64. Minimum Path Sum |
4 |
| -
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5 |
| -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. |
6 |
| -
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7 |
| -Note: You can only move either down or right at any point in time.*/ |
| 3 | +/** |
| 4 | + * 64. Minimum Path Sum |
| 5 | + * |
| 6 | + * 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. |
| 7 | + * |
| 8 | + * Note: You can only move either down or right at any point in time. |
| 9 | + */ |
8 | 10 | public class _64 {
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9 |
| - public static class Solution1 { |
10 |
| - /** |
11 |
| - * Same idea as _70: have to initialize the first row and the first column and start the for |
12 |
| - * loop from i==1 and j==1 for the rest of the matrix. |
13 |
| - */ |
14 |
| - public int minPathSum(int[][] grid) { |
15 |
| - if (grid == null || grid.length == 0) { |
16 |
| - return 0; |
17 |
| - } |
| 11 | + public static class Solution1 { |
| 12 | + /** |
| 13 | + * Same idea as _70: have to initialize the first row and the first column and start the for |
| 14 | + * loop from i==1 and j==1 for the rest of the matrix. |
| 15 | + */ |
| 16 | + public int minPathSum(int[][] grid) { |
| 17 | + if (grid == null || grid.length == 0) { |
| 18 | + return 0; |
| 19 | + } |
18 | 20 |
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19 |
| - int height = grid.length; |
20 |
| - int width = grid[0].length; |
21 |
| - int[][] dp = new int[height][width]; |
22 |
| - dp[0][0] = grid[0][0]; |
23 |
| - for (int i = 1; i < height; i++) { |
24 |
| - dp[i][0] = dp[i - 1][0] + grid[i][0]; |
25 |
| - } |
26 |
| - for (int j = 1; j < width; j++) { |
27 |
| - dp[0][j] = dp[0][j - 1] + grid[0][j]; |
28 |
| - } |
29 |
| - for (int i = 1; i < height; i++) { |
30 |
| - for (int j = 1; j < width; j++) { |
31 |
| - dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; |
| 21 | + int height = grid.length; |
| 22 | + int width = grid[0].length; |
| 23 | + int[][] dp = new int[height][width]; |
| 24 | + dp[0][0] = grid[0][0]; |
| 25 | + for (int i = 1; i < height; i++) { |
| 26 | + dp[i][0] = dp[i - 1][0] + grid[i][0]; |
| 27 | + } |
| 28 | + for (int j = 1; j < width; j++) { |
| 29 | + dp[0][j] = dp[0][j - 1] + grid[0][j]; |
| 30 | + } |
| 31 | + for (int i = 1; i < height; i++) { |
| 32 | + for (int j = 1; j < width; j++) { |
| 33 | + dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; |
| 34 | + } |
| 35 | + } |
| 36 | + return dp[height - 1][width - 1]; |
32 | 37 | }
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33 |
| - } |
34 |
| - return dp[height - 1][width - 1]; |
35 | 38 | }
|
36 |
| - } |
37 | 39 | }
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