[英]How to print the maximum valued path in a 2D array in Java?
我猜你們都知道某些人在求職面試中給您的“草莓”問題,您需要計算只能向上或向右移動的二維數組的兩個角之間的路徑,並且可以計算出最大值路徑。 我有一個完美的代碼可以在遞歸中完成,但是它的復雜性很高。 我還以O(n ^ 2)復雜度解決了“ for循環”解決方案中的問題。 但是在這種解決方案中,我只是找不到像在遞歸解決方案中那樣打印路線的方法。 這是我的代碼(在這里閱讀很長,所以我想您應該復制,編譯和運行)。 查看遞歸解決方案的結果,順便說一句-路徑需要從左下角到右上角, 我想在更好的解決方案中以相同的方式打印路線 :
public class Alg
{
public static void main(String args[])
{
String[] route = new String[100];
int[][]array = {{4,-2,3,6}
,{9,10,-4,1}
,{-1,2,1,4}
,{0,3,7,-3}};
String[][] route2 = new String[array.length][array[0].length];
int max = recursionAlg(array,array.length-1,0,route);
int max2 = loopAlg(array,array.length-1,0,route2);
System.out.println("The max food in the recursion solution is: "+max);
System.out.println("and the route is: ");
printRouteArray(route);
System.out.println("The max food in the loop solution: "+max2);
System.out.println("The route is: ");
//SHOULD PRINT HERE THE ROUTE
}
public static int loopAlg(int [][] arr,int x, int y, String[][] route)
{
int n=0;
int[][]count = new int[arr.length][arr[0].length];
for(int i = x; i>=0 ; i--)
{
for(int j = 0; j<arr[0].length; j++)
{
if (i==x && j==0) {count[i][j]=arr[i][j];}
else if (i == x) { count[i][j]=count[i][j-1]+arr[i][j];}
else if (j == 0) { count[i][j]=count[i+1][j]+arr[i][j]; }
else{
if (count[i][j-1]>count[i+1][j]) {count[i][j]=count[i][j-1]+arr[i][j];}
else { count[i][j]= count[i+1][j]+arr[i][j];}
}
}
}
return count[0][arr[0].length-1];
}
public static int recursionAlg(int [][] arr, int x, int y,String[] route)
{
return recursionAlg(arr,0,x,y,arr[0].length-1,route,0);
}
public static int recursionAlg(int[][]arr,int count,int x, int y, int max_y, String[] route, int i)
{
if (x == 0 && y == max_y) {return count;}
else if (x == 0) {
route[i]="Right";
return recursionAlg(arr,count+arr[x][y+1],x,y+1,max_y,route,i+1);
}
else if (y==max_y){
route[i]="Up";
return recursionAlg(arr,count+arr[x-1][y],x-1,y,max_y,route,i+1);
}
else if (recursionAlg(arr,count+arr[x-1][y],x-1,y,max_y,route,i+1)>recursionAlg(arr,count+arr[x][y+1],x,y+1,max_y,route,i+1))
{
route[i]="Up";
return recursionAlg(arr,count+arr[x-1][y],x-1,y,max_y,route,i+1);
}
else
{
route[i]="Right";
return recursionAlg(arr,count+arr[x][y+1],x,y+1,max_y,route,i+1);
}
}
public static void printRouteArray(String[] arr)
{
int i=0;
while (i<arr.length && (arr[i]=="Up" || arr[i]=="Right"))
{
System.out.print(arr[i]+"-->");
i++;
}
System.out.println("End");
}
}
希望能對您有所幫助,謝謝!
您需要在loopAlg
中使用另一個二維數組,該數組會記住初始2維數組中每個條目的下一步。 請參閱以下代碼和https://ideone.com/kM8BAZ進行演示:
public static void main(String args[])
{
String[] route = new String[100];
int[][]array = {{4,-2,3,6}
,{9,10,-4,1}
,{-1,2,1,4}
,{0,3,7,-3}};
String[] route2 = new String[100];
int max = recursionAlg(array,array.length-1,0,route);
int max2 = loopAlg(array,array.length-1,0,route2);
System.out.println("The max food in the recursion solution is: "+max);
System.out.println("and the route is: ");
printRouteArray(route);
System.out.println("The max food in the loop solution: "+max2);
System.out.println("The route is: ");
printRouteArray(route2);
}
public enum Dirs {START, FROM_LEFT, FROM_DOWN};
public static int loopAlg(int [][] arr,int x, int y, String[] route)
{
int n=0;
int[][]count = new int[arr.length][arr[0].length];
Dirs[][] directions = new Dirs[arr.length][arr[0].length];
List<String> path = new ArrayList<String>();
for(int i = x; i>=0 ; i--)
{
for(int j = 0; j<arr[0].length; j++)
{
if (i==x && j==0) {count[i][j]=arr[i][j]; directions[i][j] = Dirs.START;}
else if (i == x) { count[i][j]=count[i][j-1]+arr[i][j]; directions[i][j] = Dirs.FROM_LEFT;}
else if (j == 0) { count[i][j]=count[i+1][j]+arr[i][j]; directions[i][j] = Dirs.FROM_DOWN;}
else{
if (count[i][j-1]>count[i+1][j]) {count[i][j]=count[i][j-1]+arr[i][j];directions[i][j] = Dirs.FROM_LEFT;}
else { count[i][j]= count[i+1][j]+arr[i][j];directions[i][j] = Dirs.FROM_DOWN;}
}
}
}
int i=0, j=arr[0].length-1;
while(directions[i][j]!= Dirs.START) {
if(directions[i][j] == Dirs.FROM_LEFT) {
path.add("Right");
j--;
}
else {
path.add("Up");
i++;
}
}
Collections.reverse(path);
i=0;
for(String part:path) {
route[i] = part;
i++;
}
return count[0][arr[0].length-1];
}
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