一天一学: Leetcode - 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.
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.
O(1) Space: reuse input
I think it's a plus point if you mention that this is a way that the problem could be solved, and ask if you could/should do it this way. It shows that you can recognize optimization possibilities and communicative.
public int minPathSum(int[][] grid) {
int m = grid[0].length;
int n = grid.length;
for(int i=1; i<n ; i++) grid[i][0]+=grid[i-1][0];
for(int j=1; j<m ; j++) grid[0][j]+=grid[0][j-1];
for(int i=1; i<n ; i++){
for(int j=1; j<m ; j++){
grid[i][j]+=Math.min(grid[i-1][j],grid[i][j-1]);
}
}
return grid[n-1][m-1];
}
public int minPathSum(int[][] grid) {
int m = grid.length;// row
int n = grid[0].length; // column
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j != 0) {
grid[i][j] = grid[i][j] + grid[i][j - 1];
} else if (i != 0 && j == 0) {
grid[i][j] = grid[i][j] + grid[i - 1][j];
} else if (i == 0 && j == 0) {
grid[i][j] = grid[i][j];
} else {
grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j])
+ grid[i][j];
}
}
}
return grid[m - 1][n - 1];
}
public int minPathSum(int[][] grid) {
if(grid.length == 0) return 0;
int r = grid.length;
int c = grid[0].length;
for(int i=0;i<r; i++) {
for(int j=0; j<c; j++) {
int leftSum = (j>0) ? grid[i][j-1] : Integer.MAX_VALUE;
int topSum = (i>0) ? grid[i-1][j] : Integer.MAX_VALUE;
if(i==0 && j==0) continue;
grid[i][j] += Math.min(leftSum, topSum);
}
}
return grid[r-1][c-1];
}
Code from http://www.darrensunny.me/leetcode-minimum-path-sum/
// DP, O(n^2) time, O(n^2) space
Code from http://joycelearning.blogspot.com/2013/10/leetcode-minimum-path-sum.html
http://www.jiuzhang.com/solutions/minimum-path-sum/
Read full article from 一天一学: Leetcode - Minimum Path Sum
Code from http://www.darrensunny.me/leetcode-minimum-path-sum/
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0)
return 0;
int m = grid.length, n = grid[0].length;
int[] dp = new int[n];
// Find the minimum sum of all numbers along a path to grid[i][j]
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) // The first cell
dp[j] = grid[i][j];
else if (i == 0) // Cells in the first row
dp[j] = dp[j-1] + grid[i][j];
else if (j == 0) // Cells in the first column
dp[j] += grid[i][j];
else // Others
dp[j] = Math.min(dp[j-1], dp[j]) + grid[i][j];
}
}
return dp[n-1];
}
// DP, O(n^2) time, O(n^2) space
public int minPathSum(int[][] grid) { int row = grid.length; int col = grid[0].length; int[][] res = new int[row][col]; // init res[0][0] = grid[0][0]; // left for(int i = 1; i < row; i++) { res[i][0] = res[i - 1][0] + grid[i][0]; } // top for(int j = 1; j < col; j++) { res[0][j] = res[0][j - 1] + grid[0][j]; } // rest elements for(int i = 1; i < row; i++) { for(int j = 1; j < col; j++) { res[i][j] = grid[i][j] + Math.min(res[i - 1][j], res[i][j - 1]); } } return res[row - 1][col - 1]; }public int minPathSum(int[][] grid) { int row = grid.length; int col = grid[0].length; int[] res = new int[col]; // init Arrays.fill(res, Integer.MAX_VALUE); res[0] = 0; // rest elements for(int i = 0; i < row; i++) { // init the 0th sum = old 0th element + the new 0th element // just init the 0th column every time dynamically res[0] = res[0] + grid[i][0]; // loop through each element of each row for(int j = 1; j < col; j++) { res[j] = grid[i][j] + Math.min(res[j], res[j - 1]); } } return res[col - 1]; }public int minPathSum(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) { return 0; } int M = grid.length; int N = grid[0].length; int[][] sum = new int[M][N]; sum[0][0] = grid[0][0]; for (int i = 1; i < M; i++) { sum[i][0] = sum[i - 1][0] + grid[i][0]; } for (int i = 1; i < N; i++) { sum[0][i] = sum[0][i - 1] + grid[0][i]; } for (int i = 1; i < M; i++) { for (int j = 1; j < N; j++) { sum[i][j] = Math.min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j]; } } return sum[M - 1][N - 1]; }http://www.programcreek.com/2014/05/leetcode-minimum-path-sum-java/
A native solution would be depth-first search. It's time is too expensive and fails the online judgement.
