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
  
   > triangle;
	private int N;
	private Map
   
     records; public int minimumTotal(ArrayList
    
     > triangle) { this.records = new HashMap
     
      (); 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
  
   > triangle) {
		int N = triangle.size();
		Integer[] result = (Integer[]) triangle.get(N - 1).toArray(
				new Integer[N]);
		ArrayList
   
     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]; } }