【DP】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.
解法：D[i][j]保存（0,0）到（i,j）的最小和，对边缘元素要判断

public class Solution {
public int minPathSum(int[][] grid) {
int len1 = grid.length;
if(len1 == 0) return 0;
int len2 = grid[0].length;
if(len2 == 0) return 0;

int d[][] = new int[len1][len2];

for(int i=0; i<len1; i++){
for(int j=0; j<len2; j++){
d[i][j] = grid[i][j];
if(i==0 && j==0) continue;
if(i==0) d[i][j] += d[i][j-1];
else if(j==0) d[i][j] += d[i-1][j];
else{
if(d[i][j-1] < d[i-1][j]) d[i][j] += d[i][j-1];
else d[i][j] += d[i-1][j];
}
}
}
return d[len1-1][len2-1];
}
}

改进：使用滚动数组，减少空间复杂度

public class Solution {
public int minPathSum(int[][] grid) {
int len1 = grid.length;
if(len1 == 0) return 0;
int len2 = grid[0].length;
if(len2 == 0) return 0;

int d[] = new int[len2];

for(int i=0; i<len1; i++){
for(int j=0; j<len2; j++){

if(i==0 && j==0) d[j] = grid[i][j];
else if(i==0) d[j] = d[j-1] + grid[i][j];
else if(j==0) d[j] = d[j] + grid[i][j];
else{
if(d[j-1] < d[j]) d[j] = d[j-1] + grid[i][j];
else d[j] = d[j] + grid[i][j];
}
}
}
return d[len2-1];
}
}