题目

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

很明显的dp题，从下往上走，每个位置的最小的路径长度具有无后效性。

实现

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int N=triangle.size();
        int dp[N];
        for(int j=0; j<=N-1; j++){
            dp[j] = triangle[N-1][j];
        }
        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];
    }
};

思考

空间复杂度在本题中降低到了O(n)。