title
link 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.
solution
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int n=triangle.size();
vector<vector<int>> v(n,vector<int>(n,0));
for(int i=0;i<n;i++) v[n-1][i]=triangle[n-1][i];
for(int i=n-2;i>=0;i--){
for(int j=0;j<=i;j++){
v[i][j]=triangle[i][j]+min(v[i+1][j],v[i+1][j+1]);
cout<<triangle[i][j]<<" | "<<v[i][j]<<" ";
}
cout<<endl;
}
return v[0][0];
}
};
其实可以o(n)的空间复杂度的，参考link
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int n = triangle.size();
vector<int> dp(triangle.size(), 0);
for (int i = 0; i < triangle[n-1].size(); i++) {
dp[i] = triangle[n - 1][i];
}
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];
cout << dp[j] << " ";
}
cout << endl;
}
return dp[0];
}
};