120. 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.

链接： http://leetcode.com/problems/triangle/

6/7/2017

11ms, 38%

2种方法，第二种是参考别人的，虽然第15行有arraylist.remove操作，但是运行时间居然更短。

第一种：建一个初始长度为length + 1的arraylist，初始化为0，每一层从后向前赋值，并在内循环结束后删除多余的第一个元素。

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int length = triangle.size();
        List<Integer> dp = new ArrayList<Integer>();
        for (int i = 0; i <= length; i++) {
            dp.add(0);
        }
        
        for (int i = length - 1; i >= 0; i--) {
            List<Integer> row = triangle.get(i);
            int numElem = i;
            for (int j = numElem; j >= 0; j--) {
                dp.set(j + 1, row.get(j) + Math.min(dp.get(j), dp.get(j + 1)));
            }
            dp.remove(0);
        }
        return dp.get(0);
    }
}

第二种，建一个辅助arraylist，初始值为最后一行的所有元素。内循环是每一行从头赋值，不需要管最后一个元素。

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int length = triangle.size();
        List<Integer> dp = new ArrayList<Integer>(triangle.get(length - 1));
        
        for (int i = length - 2; i >= 0; i--) {
            List<Integer> row = triangle.get(i);
            for (int j = 0; j < row.size(); j++) {
                dp.set(j, row.get(j) + Math.min(dp.get(j), dp.get(j + 1)));
            }
        }
        return dp.get(0);
    }
}

更多讨论

https://discuss.leetcode.com/category/128/triangle