題目：
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.
Example:
Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
題解：
這道題跟Unique Path系列是思路一樣的。具體思路看代碼就很清楚了
代碼如下：
public class MinimumPathSum {
public static void main(String[] args) {
/**
* 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.
*
* Example:
*
* Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path
* 1→3→1→1→1 minimizes the sum.
*/
int[][] grid = { { 1, 3, 1 }, { 1, 5, 1 }, { 4, 2, 1 } };
System.out.println(minPathSum(grid));
System.out.println(minPathSum(new int[][] { { 1, 2, 5 }, { 3, 2, 1 } }));// 6
}
public static int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0] == null || grid[0].length == 0) {
return 0;
}
int rowNum = grid.length, colNum = grid[0].length;
int[][] dp = new int[rowNum][colNum];
dp[0][0] = grid[0][0];
for (int i = 1; i < rowNum; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int j = 1; j < colNum; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < rowNum; i++) {
for (int j = 1; j < colNum; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[rowNum - 1][colNum - 1];
}
}