Triangle(DP)

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

 

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

State: f[i][j] 表示跳到第(i,j)的位置的最小路径和。

Function:f[i][j] = min (f[i - 1][j - 1], f[i - 1][j])  (j != 0, j != i)

Initializtion: f[i][0] = f[i - 1][0] + a[i][0]  f[i][i] = f[ i - 1][i - 1] + a[i][i]

Answer: min(最后一行的值）

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() == 0) {
            return 0;
        }
        
        int m = triangle.size();
        int[][] f = new int[m][m];
        f[0][0] = triangle.get(0).get(0);
        for (int i = 1; i < m; i++) {
            f[i][0] = f[i - 1][0] + triangle.get(i).get(0);
            f[i][i] = f[i - 1][i - 1] + triangle.get(i).get(i);
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < i; j++) {
                f[i][j] = Math.min(f[i - 1][j], f[i - 1][j - 1]) + triangle.get(i).get(j);
            }
        }
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < m; i++) {
            min = Math.min(min, f[m - 1][i]);
        }
        return min;
    }
}