查看“︁120.Triangle”︁的源代码

{{LeetCode id=triangle}}

=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:

<syntaxhighlight lang="java">
Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output: 11
</syntaxhighlight>
Explanation: The triangle looks like:
<syntaxhighlight lang="java">
   2
  3 4
 6 5 7
4 1 8 3
</syntaxhighlight>
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).


Example 2:

<syntaxhighlight lang="java">
Input: triangle = [[-10]]
Output: -10
</syntaxhighlight>

=Solution=
==First attampt==
It works, but not good, especially memory.

* Runtime 3 ms
* Beats 65.79%
* Memory 44.4 MB
* Beats 13.95%

<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>

== Memory Optimization==

<syntaxhighlight lang="java">
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int maxRowSize = triangle.get(triangle.size() -1).size();
        // 只需要记录上一次的最短路径
        int[] dp = new int[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) {
                    // 从左到右更新，正好计算右边的元素时，当前的值已经不需要用到了。所以可以直接更新
                    dp[j] = Math.min(dp[j], dp[j+1]) + row.get(j);
                } else {
                    dp[j] = row.get(j);
                }
            }
        }
        return dp[0];
    }
}
</syntaxhighlight>

[[Category:Algorithm]]
[[Category:Dynamic Programming]]
[[Category:LeetCode]]