[leetcode]120. Triangle

题目链接：https://leetcode.com/problems/triangle/#/description

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).

方法一：（自底向上）

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

方法二：（自顶向下）

class Solution{public:    int minimumTotal(vector<vector<int>>& triangle)    {        vector<vector<int>> minilen=vector<vector<int>>(triangle.size());        for(int i=0;i<triangle.size();i++)            minilen[i].resize(triangle[i].size());        minilen[0][0]=triangle[0][0];        for(int i=1;i<triangle.size();i++)        {            for(int j=0;j<triangle[i].size();j++)            {                if(j==0)                    minilen[i][j]=minilen[i-1][j]+triangle[i][j];                else if(j==triangle[i].size()-1)                    minilen[i][j]=minilen[i-1][j-1]+triangle[i][j];                else                    minilen[i][j]=min(minilen[i-1][j-1],minilen[i-1][j])+triangle[i][j];            }        }        int minres=INT32_MAX;        for(int i=0;i<minilen.back().size();i++)        {            if (minilen.back()[i]<minres)                minres=minilen.back()[i];        }        return minres;    }};