Dynamic programming Exercises--chapter nine algorithm fifth to sixth exercises

1.Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step of the 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 11 is (i.e., 2 + 3 + 5 + 1 = 11).

//if the final optimal path passes that point, the path from the starting point to that point must be the optimal one, but the optimal path from the starting point to the point is not necessarily the local optimal path, which may be larger than the neighboring point of the same stage . Public classSolution { Public intMinimumtotal (list<list<integer>>triangle) {        intn =triangle.size (); int[] f =New intN [n];//because each step can only walk near the number, so the longest horizontal is the same as the number of rows//Initialize         for(inti=0;i<n;i++){             for(intj=0;j<n;j++) {F[i][j]=Integer.max_value; }} f[0][0]=triangle.get (0). Get (0); //let F[i][j] represent the smallest path from the starting point to the i,j point//Minimum path and f[0][n-1]~f[n-1][n-1] minimum value         for(intI =1;i<n;i++){             for(intj=0;j<=i;j++){                if(j==0) F[i][j]=f[i-1][j]+Triangle.get (i). get (j); ElseF[i][j]=math.min (F[i-1][j],f[i-1][j-1]) +Triangle.get (i). get (j); }        }        intmin=Integer.max_value;  for(intI =0;i<n;i++){            if(f[n-1][i]<min) min=f[n-1][i]; }        returnmin; }}

Lite version: Because F[I][J] is related only to f[i-1][j] and f[i-1][j-1], which is only related to the previous line state and therefore can only request 2 lines of space, I's processing on%2 is good. If the first 2 lines are related to% 3, .... And so on

1  Public classSolution {2      Public intMinimumtotal (list<list<integer>>triangle) {3         intn =triangle.size ();4         int[] f =New int[2] [n];//because each step can only walk near the number, so the longest horizontal is the same as the number of rows5          6         //Initialize7          for(inti=0;i<n;i++){8              for(intj=0;j<n;j++){9F[i%2][j]=Integer.max_value;Ten             } One         } AF[0][0]=triangle.get (0). Get (0); -         //let F[i][j] represent the smallest path from the starting point to the i,j point -         //Minimum path and f[0][n-1]~f[n-1][n-1] minimum value the          for(intI =1;i<n;i++){ -              for(intj=0;j<=i;j++){ -                 if(j==0) -F[i%2][J]=f[(i-1)%2][j]+Triangle.get (i). get (j); +                 Else -F[i%2][J]=math.min (f[(i-1)%2][J],f[(i-1)%2][J-1]) +Triangle.get (i). get (j); +             } A         } at         intmin=Integer.max_value; -          for(intI =0;i<n;i++){ -             if(f[(n-1)%2][i]<min) -Min=f[(n-1)%2 ][i]; -         } -         returnmin; in     } -}

If the final optimal path passes that point, the path from the starting point to that point must be the optimal one, but the optimal path from the starting point to the point is not necessarily the local optimal path, which may be larger than the neighboring point of the same stage.
public class Solution {
public int minimumtotal (list<list<integer>> triangle) {
int n = triangle.size ();
Int[][] f = new Int[2][n]; Because each step can only walk near the number, so the longest horizontal is the same as the number of rows

Initialization
for (int i=0;i<n;i++) {
for (int j=0;j<n;j++) {
F[i%2][j]=integer.max_value;
}
}
F[0][0]=triangle.get (0). Get (0);
Let F[i][j] represent the smallest path from the starting point to the i,j point
Minimum path and f[0][n-1]~f[n-1][n-1] minimum value
for (int i =1;i<n;i++) {
for (int j=0;j<=i;j++) {
if (j==0)
f[i%2][j]=f[(i-1)%2][j]+triangle.get (i). get (j);
Else
F[i%2][j]=math.min (f[(i-1)%2][j],f[(i-1)%2][j-1]) +triangle.get (i). get (j);
}
}
int min=integer.max_value;
for (int i =0;i<n;i++) {
if (f[(n-1)%2][i]<min)
min=f[(n-1)%2][i];
}
return min;
}
}

Dynamic programming Exercises--chapter nine algorithm fifth to sixth exercises