Triangle LeetCode |My solution

最新推荐文章于 2023-02-25 23:03:36 发布

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最新推荐文章于 2023-02-25 23:03:36 发布

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原文链接：http://www.cnblogs.com/zsychanpin/p/7189544.html

本文介绍了一种求解给定三角形从顶点到底部的最小路径和的方法。通过动态规划思想，创建二维数组存储每一步到达各点的最小路径和，并最终找到最小路径总和。

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).

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

    }
}

转载于:https://www.cnblogs.com/zsychanpin/p/7189544.html