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
     
      
       > triangle) {        int m = triangle.size();        return minPath(triangle, 0, 0, m, 0);    }        private int minPath(List
       
        
         > 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
     
      
       > 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];    }}