[Array]Triangle

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.

方法1：从上向下进行遍历dp，但是这样写的代码比较繁琐。

class Solution {public:    int minimumTotal(vector<vector<int>>& triangle) {        if(triangle.size() == 0||triangle[0].size()==0)            return 0 ;        int* buffer = new int[triangle[triangle.size() - 1].size()];        buffer[0] = triangle[0][0];        for(int row = 1 ; row < triangle.size(); ++row){            for(int col = triangle[row].size() - 1; col >= 0; --col){                if(col == triangle[row].size() - 1){                    buffer[col] = buffer[col - 1]+ triangle[row][col];                }                 else if(col == 0){                    buffer[col] = buffer[col] + triangle[row][col];                }                else{                    buffer[col] = min(buffer[col-1],buffer[col]) + triangle[row][col];                }            }        }        int sum = INT_MAX;        for(int i = 0 ; i < triangle[triangle.size() - 1].size(); ++i){            if(sum > buffer[i])                sum = buffer[i];        }        return sum;    }};

方法2：从下向上进行遍历dp，这样的代码相对从上到下而言比较简洁。

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