## Saturday, November 5, 2016

### LeetCode OJ - Triangle

Problem:

Solution:

At any layer, if we know the cheapest path from any parent to get to the root, then we know the cheapest path to reach itself by picking the cheaper parent!

Note the use of double buffer technique to ensure $O(n)$ storage, where $n$ is the height of the triangle.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/triangle/

#include "LEET_TRIANGLE.h"
#include <algorithm>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_TRIANGLE
{
class Solution
{
public:
int minimumTotal(vector<vector<int>>& triangle)
{
int n = triangle.size();
int answer = ~(1 << 31);
vector<int> buffer1;
vector<int> buffer2;
buffer1.resize(n);
buffer2.resize(n);
vector<int>* pBuffer1 = &buffer1;
vector<int>* pBuffer2 = &buffer2;
for (int i = 0; i < n; i++)
{
vector<int>& last_minimum = *pBuffer1;
vector<int>& this_minimum = *pBuffer2;
if (i == 0)
{
this_minimum[0] = triangle[0][0];
}
else
{
for (int j = 0; j < i + 1; j++)
{
if (j == 0)
{
this_minimum[0] = last_minimum[0];
}
else if (j == i)
{
this_minimum[i] = last_minimum[i - 1];
}
else
{
this_minimum[j] = min(last_minimum[j - 1], last_minimum[j]);
}
this_minimum[j] += triangle[i][j];
}
}
if (i == n - 1)
{
for (int j = 1; j < i + 1; j++)
{
}
}

swap(pBuffer1, pBuffer2);
}
}
};
};

using namespace _LEET_TRIANGLE;

int LEET_TRIANGLE()
{
Solution solution;
vector<int> a;
vector<int> b;
vector<int> c;
vector<int> d;
vector<vector<int>> triangle;
a.push_back(2);
b.push_back(3); b.push_back(4);
c.push_back(6); c.push_back(5); c.push_back(7);
d.push_back(4); d.push_back(1); d.push_back(8); d.push_back(3);
triangle.push_back(a);
triangle.push_back(b);
triangle.push_back(c);
triangle.push_back(d);
cout << solution.minimumTotal(triangle) << endl;
return 0;
}