leetcode 64. 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.

很简单的一道题，明显的DP做法。

下面是grid和DP的一个例子：



public int minPathSum(int[][] grid) {
int m=grid.length;
int n=grid[0].length;
if(m==0||n==0){
return 0;
}
//dp[i][j]存储走到（i,j）这个点的最小sum
int[][] dp=new int[m]
;
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 i=1;i<n;i++){
dp[0][i]=dp[0][i-1]+grid[0][i];
}
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];
}
大神说可以把 dp 的空间缩减为一维：

public class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[] dp = new int[n+1];
for(int i = 0; i < n; i++){
dp[i+1] = grid[0][i] + dp[i];
}
for (int i = 1; i < m; i++) {
for (int j = 0; j < n; j++) {
if(j == 0) {
dp[j+1] = dp[j+1] + grid[i][j];
}else {
dp[j+1] = Math.min(dp[j],dp[j+1]) + grid[i][j];
}
}
}
return dp
;
}
}