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).Analysis:
This problem is more likely to be a (dynamic programming) DP problem,
where a[n][i] = a[n][i]+min(a[n-1][i], a[n-1][i-1]).
Note that in this problem, "adjacent" of a[i][j] means a[i-1][j] and a[i-1][j-1], if available(not out of bound), while a[i-1][j+1] is not "adjacent" element.
The minimum of the last line after computing is the final result.
Code:
class Solution {
public:
int minimumTotal(vector<vector<int> > &triangle) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int n = triangle.size();
if (n==0){return 0;}
for (int i=1;i<n;i++){
for (int j=0;j<triangle[i].size();j++){
if (j==0){triangle[i][j] += triangle[i-1][j];}
if (j>0){
if (i>j){
triangle[i][j] += min(triangle[i-1][j-1],triangle[i-1][j]);
}else{
triangle[i][j] += triangle[i-1][j-1];
}
}
}
}
sort(triangle[n-1].begin(),triangle[n-1].end());
return triangle[n-1][0];
}
};
int minimumTotal(vector > &triangle) {
ReplyDeletefor (int i = triangle.size() - 2; i >= 0; --i)
for (int j = 0; j < i + 1; ++j)
triangle[i][j] += min(triangle[i + 1][j], triangle[i + 1][j + 1]);
return triangle[0][0];
}
this is with o(N) space complexity logic
ReplyDeleteprivate int minimumTotal(ArrayList> a) {
int m, n;
m = a.size();
n = a.get(m - 1).size();
int [] dp = new int[n + 1];
int size = n;
for (int i = m - 1; i >= 0; i--) {
for (int j = 0; j < size; j++) {
dp[j] = Math.min(dp[j], dp[j + 1]) + a.get(i).get(j);
}
size--;
}
return dp[0];
}