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.
这道题依然是62题的followup，不过要求的结果变为了最小路径。同样用之前的思路，我们定义的二维数组dp代表的是从起点到这个位置的最短路径，第一行和第一列都只和它前面的位置有关，其他的位置等于左边或者上边最短路径的最小值加上自己这个点的值。
即递推公式是dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] （i > 0 && j > 0）。
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[m][n];
dp[0][0] = grid[0][0];
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i-1][0] + grid[i][0];
}
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j-1] + grid[0][j];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
}
}
return dp[m-1][n-1];
}