LeetCode – Minimum Path Sum (Java)
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.
Java Solution 1: Depth-First Search
A native solution would be depth-first search. It's time is too expensive and fails the online judgement.
public int minPathSum(int[][] grid) { return dfs(0,0,grid); } public int dfs(int i, int j, int[][] grid){ if(i==grid.length-1 && j==grid[0].length-1){ return grid[i][j]; } if(i<grid.length-1 && j<grid[0].length-1){ int r1 = grid[i][j] + dfs(i+1, j, grid); int r2 = grid[i][j] + dfs(i, j+1, grid); return Math.min(r1,r2); } if(i<grid.length-1){ return grid[i][j] + dfs(i+1, j, grid); } if(j<grid[0].length-1){ return grid[i][j] + dfs(i, j+1, grid); } return 0; } |
Java Solution 2: Dynamic Programming
public int minPathSum(int[][] grid) { if(grid == null || grid.length==0) return 0; int m = grid.length; int n = grid[0].length; int[][] dp = new int[m][n]; dp[0][0] = grid[0][0]; // initialize top row for(int i=1; i<n; i++){ dp[0][i] = dp[0][i-1] + grid[0][i]; } // initialize left column for(int j=1; j<m; j++){ dp[j][0] = dp[j-1][0] + grid[j][0]; } // fill up the dp table for(int i=1; i<m; i++){ for(int j=1; j<n; j++){ if(dp[i-1][j] > dp[i][j-1]){ dp[i][j] = dp[i][j-1] + grid[i][j]; }else{ dp[i][j] = dp[i-1][j] + grid[i][j]; } } } return dp[m-1][n-1]; } |
<pre><code> String foo = "bar"; </code></pre>
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