Leetcode DFS & DP triangle

Triangle total accepted: 17536 total submissions: 65508my submissions

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 is11(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.

Given a triangular array, each position in the triangular array has a numerical value.
Find a path from the top to the bottom to minimize the total number in the path.
You can go left or right of the next row at the current position each time.
Thought 1: DFS + Memorandum
In the J position of line I, there are two options to go down: the J position to line I + 1 or the J + 1 position
The subproblem at the J position or J + 1 position in line I + 1 is the same as the original problem and can be solved recursively by DFS.
Use an array _ Min [I] [J] To save the value of the solution process to avoid repeated computation.
Complexity: time O (N ^ 2), Space O (N ^ 2)


Idea 2: DP + scrolling Array
If DP [I] [J] is set to the minimum value of the path at the position J of row I, the state transition equation is
DP [I] [J] = min (DP [I-1] [k]) + A [I] [J] Where DP [I-1] [k] represents the previous state that can reach the DP [I] [J] State
Here, K can take J-1 or J, a [I] [J] to indicate the value of this position.
Because row I only needs to use the data of the row I-1, you do not have to save the optimal value before the I-1.
Finally, you only need to use the array DP [J ].
Complexity: time O (N ^ 2), Space O (N)

// Train of Thought 1 vector <int> _ min, _ triangle; int DFS (int I, Int J) {if (I = _ triangle. size ()-1) {return _ triangle [I] [J];} If (j <0 | j> = _ triangle [I]. size () return int_max; If (_ Min [I] [J]! = Int_max) return _ Min [I] [J]; // a special value indicates DFS (I, j) return _ Min [I] [J] = min (DFS (I + 1, J), DFS (I + 1, J + 1 )) + _ triangle [I] [J];} int minimumtotal (vector <int> & triangle) {If (triangle. empty () return 0; _ triangle = triangle; _ min = vector <int> (triangle. size (), vector <int> (Triangle [triangle. size ()-1]. size (), int_max); Return DFS (0, 0);} // Train of Thought 2int minimumtotal (vector <int> & triangle) {If (triangle. empty () return 0; vector <int> dp (triangle. back (). size (), 0); // initialize int I = 0; for_each (Triangle [0]. begin (), triangle [0]. end (), [& I, & DP] (Int & V) {DP [I ++] = V ;}); // iteratively update dpfor (I = 1; I <triangle. size (); ++ I) {int dp_j_1 = DP [0]; for (Int J = 0; j <triangle [I]. size (); ++ J) {int TEM = DP [J]; If (j = triangle [I]. size ()-1) DP [J] = dp_j_1; DP [J] = min (dp_j_1, DP [J]) + triangle [I] [J]; dp_j_1 = TEM ;}} int _ min = int_max; For (Int J = 0; j <DP. size (); ++ J) _ min = min (_ min, DP [J]); Return _ min ;}

Leetcode DFS & DP triangle