leetcode-triangle

虽说题目很简单，也没有优化。但是depend on myself & 一遍就AC还是很让人开心的。。

之前二维vector不熟，通过这个题熟了很多。

时间复杂度O(n^2),空间复杂度O(n^2). 优化之后可以做到空间复杂度O(1), 见代码2.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        if (triangle.empty())
            return 0;
        vector<vector<int>> d;
        for (int i = 0; i < triangle.size(); i++)
        {
            d.push_back(vector<int>(triangle[i].size(), INT_MAX));
        }
        d.at(0).at(0) = (triangle.at(0).at(0));
        for (int i = 1; i < triangle.size(); i++)
        {
            for (int j = 0; j < triangle[i].size(); j++)
            {
                if (j == 0)
                {
                    d.at(i).at(j) = triangle.at(i).at(j) + d.at(i - 1).at(j);
                }
                else if (j == int(triangle.at(i).size()) - 1)
                {
                    d.at(i).at(j) = triangle.at(i).at(j) + d.at(i - 1).at(j - 1);
                }
                else
                {
                    d.at(i).at(j) = triangle.at(i).at(j) + min(d.at(i - 1).at(j - 1), d.at(i - 1).at(j));
                }
            }
        }
        int minv = d.at(int(d.size()) - 1).at(0);
        for (int i = 1; i < int(d.at(int(d.size())-1).size()); i++)
        {
            if (d.at(d.size()-1).at(i)<minv)
            {
                minv = d.at(d.size() - 1).at(i);
            }
        }
        return minv;
    }
};

int main()
{
    Solution s;
    vector<vector<int>> tri;
    int a1[] = { 2 };
    int a2[] = { 3, 4 };
    int a3[] = { 6, 5, 7 };
    int a4[] = { 4, 1, 8, 3 };
    //tri.push_back(vector<int>(a1,a1+1));
    //tri[0].push_back(1); tri[0].push_back(9);
    tri.push_back(vector<int>(a1, a1 + 1));
    tri.push_back(vector<int>(a2, a2 + 2));
    tri.push_back(vector<int>(a3, a3 + 3));
    tri.push_back(vector<int>(a4, a4 + 4));
    cout<<s.minimumTotal(tri)<<endl;
    return 0;
}

 以下为优化后：空间复杂度变为O(1)。

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        if (triangle.empty())
            return 0;
        for (int i = triangle.size() - 2; i >= 0; i--)
        {
            for (int j = 0; j < i + 1; j++)
            {
                triangle[i][j] += min(triangle[i+1][j],triangle[i+1][j+1]);
            }
        }
        return triangle[0][0];
    }
};