Leetcode: Triangle

最新推荐文章于 2019-07-27 09:27:00 发布

原创最新推荐文章于 2019-07-27 09:27:00 发布 · 579 阅读

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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。通过使用仅需O(n)额外空间的高效算法实现，其中n为三角形总行数。示例中展示了如何对特定三角形数据进行操作并找到最小路径。

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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.

public class Solution {
    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
		// Start typing your Java solution below
		// DO NOT write main() function
		int[] sum = new int[triangle.size()];
		if (triangle.size() < 1)
			return 0;
		int min = sum[0], pre_sum = 0;
		for (int i = 0; i < triangle.size(); i++) {
			for (int j = 0; j <= i; j++) {
				if (j == 0){
					pre_sum = sum[j];
					sum[j] = sum[j] + triangle.get(i).get(j);
					min = sum[0];
				}
				else if(j == i)
					sum[j] = pre_sum + triangle.get(i).get(j);
				else {
					int temp = pre_sum;
					pre_sum = sum[j];
					sum[j] = temp < sum[j] ?
							temp + triangle.get(i).get(j)
							: sum[j] + triangle.get(i).get(j);
				}
				if (sum[j] < min)
					min = sum[j];
			}
		}
		return min;
	}
}