[leetcode-120]Triangle(c++)

问题描述：
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.

分析：使用动态规划算法，对于每一行而言，f(i,j) = min(f(i-1,j-1),f(i-1,j)+val; 然后查看最后一行的情况。

代码如下：8ms

class Solution {public:    int minimumTotal(vector<vector<int>>& triangle) {        int n = triangle.size();        if(n<=0)            return 0;        vector<int> prev = triangle.at(0);        if(prev.size()<=0)            return 0;        for(int i = 1;i<n;i++){            vector<int> cur = triangle.at(i);            for(int j = 0;j<cur.size();j++){                int val = INT_MAX;                if(j-1>=0){                    if(val>prev.at(j-1))                        val = prev.at(j-1);                }                if(j<prev.size())                    if(val>prev.at(j))                        val = prev.at(j);                cur[j] += val;            }            prev = cur;        }        int min = prev.at(0);        for(int i = 1;i<prev.size();i++){            if(min>prev.at(i))                min = prev.at(i);        }        return min;    }};