Triangle

投稿日 |

Description

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

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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.

Method

这是一个动态规划问题，用一个数组存入每一个数字的最优值，然后求最后一行的最小值即可。
具体流程如下：

  1. 初始dp[0][0] = triangle.get(0).get(0)。
  2. dp[i][j]会等于相邻父亲数字里最小值加上他本身的值。如果是端点则直接等于父亲数字的最优值加上它本身。
  3. 求最后一行的最小值。

Solution

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