LeetCode——Triangle

最新推荐文章于 2024-01-06 17:59:43 发布

Paul_ZJ

最新推荐文章于 2024-01-06 17:59:43 发布

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分类专栏： LeetCode——坚持每天一题

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本文介绍了一种寻找从三角形顶部到底部的最小路径和的方法，通过动态规划算法实现，确保了高效求解。

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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.

Java代码：

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int total = 0;
		int[][] sum = new int[triangle.size()][];
		for (int i = 0; i < triangle.size(); i++)
			sum[i] = new int[triangle.get(i).size()];
		sum[0][0] = triangle.get(0).get(0);
		for (int i = 1; i < triangle.size(); i++)
			sum[i][0] = sum[i - 1][0] + (int) triangle.get(i).get(0);

		for (int i = 1; i < triangle.size(); i++)
			sum[i][triangle.get(i).size() - 1] = sum[i - 1][triangle.get(i - 1)
					.size() - 1]
					+ triangle.get(i).get(triangle.get(i).size() - 1);

		for (int i = 2; i < triangle.size(); i++)
			for (int j = 1; j < triangle.get(i).size() - 1; j++)
				if (sum[i - 1][j - 1] > sum[i - 1][j])
					sum[i][j] = sum[i - 1][j] + triangle.get(i).get(j);
				else
					sum[i][j] = sum[i - 1][j-1] + triangle.get(i).get(j);

		total = sum[triangle.size() - 1][0];
		for (int i = 1; i < triangle.get(triangle.size() - 1).size(); i++)
			if (total > sum[triangle.size() - 1][i])
				total = sum[triangle.size() - 1][i];

		return total;
    }
}