Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
Example
Given the following triangle
1
2
3
4
5
6
| [
[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.
Solution
(Triangle.java) download
1
2
3
4
5
6
7
8
9
10
11
12
13
14
| public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int n = triangle.size();
int[] dp = new int[n];
for (int i=0;i<n;i++)
dp[i] = triangle.get(n-1).get(i);
for (int i=n-2; i>=0; i--){
for (int j=0;j<=i;j++){
dp[j] = Math.min(dp[j+1],dp[j])+triangle.get(i).get(j);
}
}
return dp[0];
}
}
|
(Triangle.py) download
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| class Solution(object):
def minimumTotal(self, triangle):
"""
:type triangle: List[List[int]]
:rtype: int
"""
#dp[i][j]: 某一行第i列 = List[i][j] + min(dp[i-1][j-1], dp[i-1][j+1])
#记住上一行的所有列的dp即可
n = len(triangle)
dp = [0]*n
#init: from bottom to top
for i in range(n):
dp[i] = triangle[n-1][i]
for i in range(n-2,-1,-1):
for j in range(i+1):
dp[j] = triangle[i][j]+min(dp[j],dp[j+1])
return dp[0]
|