LeetCode | Triangle

题目

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.

有从上到下递归的解法，空间复杂度过高：为了防止超时，需要记录各个节点的最小值，同时不断的压栈也是空间的消耗。见解法1

这题最适合的从下到上直接求解，直接根据下一行计算当前行的最小值即可。由题目可知，不希望你更改原有的数据，额外开辟的O(n)数组纪录就是了。见解法2

这题没写入参是否为null的判断了- - ！

解法1

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;

public class Triangle {
	private ArrayList<ArrayList<Integer>> triangle;
	private int N;
	private Map<Integer, Integer> records;

	public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
		this.records = new HashMap<Integer, Integer>();
		this.triangle = triangle;
		this.N = triangle.size();
		return solve(0, 0);
	}

	private int solve(int level, int index) {
		int key = level * N + index;
		if (records.containsKey(key)) {
			return records.get(key);
		}
		int value = triangle.get(level).get(index);
		if (level == N - 1) {
			records.put(key, value);
			return value;
		}
		value += Math.min(solve(level + 1, index), solve(level + 1, index + 1));
		records.put(key, value);
		return value;
	}
}

public class Triangle {
	public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
		int N = triangle.size();
		Integer[] result = (Integer[]) triangle.get(N - 1).toArray(
				new Integer[N]);
		ArrayList<Integer> currentRow = null;
		for (int i = N - 2; i >= 0; --i) {
			currentRow = triangle.get(i);
			for (int j = 0; j < i + 1; ++j) {
				result[j] = Math.min(result[j], result[j + 1])
						+ currentRow.get(j);
			}
		}
		return result[0];
	}
}

LeetCode | Triangle

原文：http://blog.csdn.net/perfect8886/article/details/19853101