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.

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
Read full article from 一天一学: Leetcode - Minimum Path Sum
Note: You can only move either down or right at any point in time.

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]; }
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