LeetCode

题目：

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).

Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

思路：

1) 递归，代码很简单，但超时了

package recursion;

import java.util.ArrayList;
import java.util.List;

public class Triangle {

    public int minimumTotal(List<List<Integer>> triangle) {
        int m = triangle.size();
        return minPath(triangle, 0, 0, m, 0);
    }
    
    private int minPath(List<List<Integer>> triangle, int row, int col, int m, int prevTotal) {
        if (row >= m) return prevTotal;
        prevTotal += triangle.get(row).get(col);
        return Math.min(minPath(triangle, row + 1, col, m, prevTotal), minPath(triangle, row + 1, col + 1, m, prevTotal));
    }

}

2) 从下往上进行扫描

package recursion;

import java.util.ArrayList;
import java.util.List;

public class Triangle {

    public int minimumTotal(List<List<Integer>> triangle) {
        int m = triangle.size();
        int n = triangle.get(m - 1).size();
        int[] res = new int[n + 1];
        for (int i = m - 1; i >= 0; --i) {
            for (int j = 0; j < triangle.get(i).size(); ++j) {
                res[j] = Math.min(res[j], res[j + 1]) + triangle.get(i).get(j);
            }
        }
        return res[0];
    }

}