Minimum Path Sum

https://oj.leetcode.com/problems/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.

解题思路：

典型的动态规划。将状态dp[m]
定义为，到该点的sum。dp[m]
= min{dp[m - 1]
, dp[m][n - 1]} + grid[m]
。

public class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;

int[][] sum = new int[m]
;
sum[0][0] = grid[0][0];
for(int i = 1; i < n; i++){
sum[0][i] = sum[0][i - 1] + grid[0][i];
}

for(int i = 1; i < m; i++){
sum[i][0] = sum[i - 1][0] + grid[i][0];
}

for(int i = 1; i < m; i++){
for(int j = 1; j < n; j++){
sum[i][j] = Math.min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
}
}
return sum[m - 1][n - 1];
}
}

与上一题一样，仍然可以用一个一维数组定义这个dp，以节省空间。sum[j] = Math.min(sum[j], sum[j - 1]) + grid[i][j];sum永远保存本行的sum[j]，所以到点[i][j]，前一个点的dp[i - 1][j]已更新，比较他和当前dp[j]（未更新，代表上一行）的值，然后再更新当前dp[j]。

public class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;

int[] sum = new int
;
sum[0] = grid[0][0];

for(int i = 1; i < n; i++){
sum[i] = sum[i - 1] + grid[0][i];
}

for(int i = 1; i < m; i++){
for(int j = 0; j < n; j++){
if(j ==0){
sum[j] = sum[j] + grid[i][j];
}else {
sum[j] = Math.min(sum[j], sum[j - 1]) + grid[i][j];
}
}
}
return sum[n - 1];
}
}