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]; } }