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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 naive solution would be a depth-first search. Its time is too expensive and fails online judgment.
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];
}