[leetcode] Triangle

最新推荐文章于 2020-05-23 10:06:18 发布

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本文介绍了一个算法问题——给定一个三角形结构的数组，寻找从顶点到底部的最小路径总和。提供了两种解决方案，分别使用C++和Java实现，详细展示了如何通过动态规划来高效解决这个问题。

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From : https://leetcode.com/problems/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).

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if(triangle.size() == 0 || triangle[0].size() == 0) return 0;
        vector<int> mins;
        mins.push_back(triangle[0][0]);
        for(int i=1, rows=triangle.size(); i<rows; i++) {
            mins.push_back(triangle[i][i]+mins[i-1]);
            for(int j=i-1; j>0; j--) {
                mins[j] = min(mins[j], mins[j-1])+triangle[i][j];
            }
            mins[0] += triangle[i][0];
        }
        int min = INT_MAX;
        for(int i=mins.size()-1; i>=0; i--) {
            if(min > mins[i]) min = mins[i];
        }
        return (min!=INT_MAX)*min;
    }
};

public class Solution {
	public int minimumTotal(List<List<Integer>> t) {
		if (t == null || t.size() == 0 || t.get(0) == null
				|| t.get(0).size() == 0) {
			return 0;
		}
		int[] candidates = new int[t.size()];
		candidates[0] = t.get(0).get(0);
		for (int i = 1; i < t.size(); ++i) {
			List<Integer> cur = t.get(i);
			for (int j = i; j >= 0; --j) {
				if (j == 0 || j == i) {
					if (j == 0) {
						candidates[0] += cur.get(0);
					} else {
						candidates[j] = candidates[j - 1] + cur.get(j);
					}
				} else {
					candidates[j] = cur.get(j)
							+ Math.min(candidates[j], candidates[j - 1]);
				}
			}
		}
		int ans = candidates[0];
		for (int i = 1; i < candidates.length; ++i) {
			if (candidates[i] < ans) {
				ans = candidates[i];
			}
		}
		return ans;
	}
}