[leetcode] 120. 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.

解法一：

逐层叠加。非O(n) space。

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if(!triangle.size()) return 0;
        vector<int> sum(1,triangle[0][0]);
        
        for(int i= 1; i < triangle.size(); i++){
            vector<int> tmp(triangle[i].size(),0);
            for(int j = 0; j<triangle[i].size(); j++){
                tmp[j] = min(sum[max(0,j-1)],sum[min(int(sum.size()-1),j)]) + triangle[i][j];
            }
            sum = tmp;
        }
        sort(sum.begin(),sum.end());
        return sum[0];
    }
};

解法二：

思想类似，只不过从最后一层网上做dp。

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