[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.

 

Solution:

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null)
            return 0;
        int length = triangle.size();
        int dp[] = new int[length];
        for (int i = 0; i < length; ++i) {
            dp[i] = triangle.get(length - 1).get(i);
        }
        for (int j = length - 2; j >= 0; --j) {
            int[] temp = dp.clone();
            for (int i = j; i >= 0; --i) {
                temp[i] = Math.min(dp[i], dp[i + 1]) + triangle.get(j).get(i);
            }
            dp = temp.clone();
        }
        return dp[0];
    }
}

rewrite:

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

 前边儿的代码，在每一层上从后往前算，需要再开一个temp数组来放下一层的dp值；

rewrite后的代码，直接在dp上进行操作，节省空间和时间。