【LeetCode】120. Triangle

Difficulty: Medium

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.

f[i][j]表示从（0,0）走到（i,j）的最小和

状态转移方程：

f[i][0]=f[i-1][0]+triangle[i][0]

f[i][j]=min(f[i-1],[j-1],f[i-1][j]) + triangle[i][j]

f[i][i]=f[i-1][i-1] + triangle[i][i]

时间复杂度O(n * n)

实现O(n)
extra space只需要用滚动数组就可以了，见代码

class Solution {
public:
int minimumTotal(vector<vector<int> >& triangle) {
int i,j,k,n;
n=triangle.size();
vector<int> f[2];
f[0].resize(n);
f[1].resize(n);
f[0][0]=triangle[0][0];

for(i=k=1;i<n;i++,k=1-k)
{
f[k][0]=f[1-k][0] + triangle[i][0];
f[k][i]=f[1-k][i-1] + triangle[i][i];
for(j=1;j<i;j++)
f[k][j]=min(f[1-k][j-1],f[1-k][j])+triangle[i][j];
}

int res=f[1-k][0];
for(i=1;i<n;i++)
res=min(res,f[1-k][i]);
return res;
}
};