*Triangle

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.

题解：

一道动态规划的经典题目。需要自底向上求解。

递推公式是： dp[i][j] = dp[i+1][j] + dp[i+1][j+1] ，当前这个点的最小值，由他下面那一行临近的2个点的最小值与当前点的值相加得到。

由于是三角形，且历史数据只在计算最小值时应用一次，所以无需建立二维数组，每次更新1维数组值，最后那个值里存的就是最终结果。

代码如下：

public int minimumTotal(List<List<Integer>> triangle) {
    if(triangle.size()==1)
        return triangle.get(0).get(0);
        
    int[] dp = new int[triangle.size()];

    //initial by last row 
    for (int i = 0; i < triangle.get(triangle.size() - 1).size(); i++) {
        dp[i] = triangle.get(triangle.size() - 1).get(i);
    }
 
    // iterate from last second row
    for (int i = triangle.size() - 2; i >= 0; i--) {
        for (int j = 0; j < triangle.get(i).size(); j++) {
            dp[j] = Math.min(dp[j], dp[j + 1]) + triangle.get(i).get(j);
        }
    }
 
    return dp[0];
}

Reference:  http://www.programcreek.com/2013/01/leetcode-triangle-java/