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)。
