120.Triangle:修订间差异

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==Solution==
==Solution==
===First attampt==


<syntaxhighlight lang="java">
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int maxRowSize = triangle.get(triangle.size() -1).size();
        // create an array nxn to store the minimal distance that starts from triangle[i][j]
        int[][] dp = new int[maxRowSize][maxRowSize];
        for(int i = triangle.size() -1; i >=0; i--) {
            List<Integer> row = triangle.get(i);
            for(int j = 0; j < row.size(); j++) {
                if(i < dp.length -1) {
                    // calculate minimal distance and update
                    dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + row.get(j);
                } else {
                    // for the bottom line, there's no subpaths
                    dp[i][j] = row.get(j);
                }
            }
        }
        return dp[0][0];
    }
}
</syntaxhighlight>





2023年9月23日 (六) 04:01的版本

Description

https://leetcode.com/problems/triangle/

Given a triangle array, return the minimum path sum from top to bottom.


For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.


Example 1:

Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output: 11

Explanation: The triangle looks like:

   2
  3 4
 6 5 7
4 1 8 3

The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).


Example 2:

Input: triangle = [[-10]]
Output: -10

Solution

=First attampt

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int maxRowSize = triangle.get(triangle.size() -1).size();
        // create an array nxn to store the minimal distance that starts from triangle[i][j]
        int[][] dp = new int[maxRowSize][maxRowSize];

        for(int i = triangle.size() -1; i >=0; i--) {
            List<Integer> row = triangle.get(i);
            for(int j = 0; j < row.size(); j++) {
                if(i < dp.length -1) {
                    // calculate minimal distance and update
                    dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + row.get(j);
                } else {
                    // for the bottom line, there's no subpaths
                    dp[i][j] = row.get(j);
                }
            }
        }
        return dp[0][0];
    }
}