Triangle

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.

Ref:http://www.cnblogs.com/feiling/p/3269609.html

[解题思路]

该题是经典的DP问题，状态可以定义成dp[node]表示从当前node到bottom的最小路径和，对于最下面一层，因为它们是最底层，故它们到bottom的最小路径和就是它们自身；再往上一层，如节点6，它到达bottom的最小路径和即为节点4与节点1之间的最小值加上节点6自身的值

由以上分析得出状态迁移方程：

dp[node] = value[node] + min(dp[child1], dp[child2])

另外本题要求时间复杂度为：O(n), n为三角形的行数

public class Solution {
    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
        if(triangle == null || triangle.size() == 0){
            return 0;
        }
        int row = triangle.size();
        int[] num = new int[row];
        for(int i = row-1; i>= 0; i--){
            int col = triangle.get(i).size();
            for(int j = 0; j< col; j++){
                if(i == row-1){
                    num[j] = triangle.get(i).get(j);
                    continue;
                }
                num[j] = Math.min(num[j] , num[j+1])+ triangle.get(i).get(j);
            }
        }
     return num[0];   
    }
}

Triangle

原文：http://www.cnblogs.com/RazerLu/p/3552577.html