[Leetcode] 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.

比较简单的DP题目，状态转移方程为f(i, j) = f(i-1, j-1) + f(i-1, j) + c(i, j). 最直接的方法是是用二维数组去存计算的结果，但是我们发现每一次的dp只用到了上一次的dp结果，因此实际使用一个一维数组就足够了，注意更新的时候从后面往前更新，因为我们需要用到之前的结果。

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