我寫了這個工作解決方案,我分享給任何感興趣的人,但是當涉及到處理大數組(例如3000)時,會拋出java堆空間錯誤,因爲代碼將消耗大量內存,或建議可以理解
public class Solution {
static int startPosition;
public static int compute(BufferedReader br) throws IOException {
final int totalNodeCount = getTotalNodeNumber(br);
final String caseArray[] = new String[totalNodeCount];
bufferToArray(br, caseArray);
startPosition = getStartPosition(caseArray);
final boolean visited[] = new boolean[caseArray.length];
int minimumNumberOfMove = 0;
final List<Integer> reachableList = new ArrayList<Integer>();
for (int i = 1; i <= 6; i++)
{
visitedInitilise(visited);
if (((startPosition + i) < totalNodeCount) && ((startPosition + i) > 0))
getMinimumNumberOfMoves(caseArray, visited, startPosition + i, 0, reachableList);
}
// Retriving Minimum number of move from all reachble route
if (reachableList.isEmpty())
minimumNumberOfMove = Constants.IMPOSSIBLE;
else
{
minimumNumberOfMove = reachableList.get(0);
for (int i = 0; i < reachableList.size(); i++)
if (reachableList.get(i) < minimumNumberOfMove)
minimumNumberOfMove = reachableList.get(i);
}
return minimumNumberOfMove;
}
static int getStartPosition(String[] plateau){
int startIndex = 0;
for (int i = 0; i <= (plateau.length - 1); i++)
if (plateau[i].equals("S"))
{
startIndex = i;
break;
}
return startIndex;
}
static void bufferToArray(BufferedReader br, String[] plateau) {
String line;
int i = 0;
try
{
while ((line = br.readLine()) != null)
{
plateau[i] = line;
i++;
}
}
catch (IOException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
}
static int getTotalNodeNumber(BufferedReader br) {
int i = 0;
try
{
i = Integer.parseInt(br.readLine());
} catch (NumberFormatException e)
{
e.printStackTrace();
} catch (IOException e)
{
e.printStackTrace();
}
return i;
}
static List<Integer> getMinimumNumberOfMoves(String[] plateau, boolean[] visited, final int currentIndex,
int currentNumberOfMoves, List<Integer> list) {
Boolean endIsReached = false;
Boolean impossible = false;
visited[startPosition] = true;
// Checking if the current index index is negativ
if (currentIndex < 0)
impossible = true;
while ((endIsReached == false) && (impossible == false) && (visited[currentIndex] == false)
&& (currentIndex < plateau.length))
{
try
{
switch (plateau[currentIndex]) {
case "E": {
// if end is reached , pushing number of move into our list
endIsReached = true;
list.add(currentNumberOfMoves + 1);
break;
}
case "R": {
// Marking node as visited
visited[currentIndex] = true;
for (int i = 1; i <= 6; i++)
{
// Marking all case after R case as non visited
for (int j = currentIndex + 1; j < visited.length; j++)
visited[j] = false;
// Calculating number of move after R case
if (((currentIndex + i) < plateau.length) && (currentIndex > 0))
getMinimumNumberOfMoves(plateau, visited, currentIndex + i, currentNumberOfMoves + 1, list);
}
break;
}
default: {
// Cheking if node was already visited
if (visited[currentIndex] == true)
{
// Marking all node as non visited
visitedInitilise(visited);
impossible = true;
break;
}
else
{
// when the node was not visited before , catch the jump
// value
int jumpValue = Integer.parseInt(plateau[currentIndex]);
// cheking that the next node is not bigger than node
// number and not negativ
if (((currentIndex + jumpValue) > plateau.length) || (currentIndex < 0))
{
impossible = true;
break;
}
else
{
// Marking node as visited
visited[currentIndex] = true;
// calculating minimum number of move starting from
// this node
getMinimumNumberOfMoves(plateau, visited, currentIndex + jumpValue,
currentNumberOfMoves + 1, list);
break;
}
}
}
}
}
catch (NumberFormatException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
break;
}
if (impossible == true)
currentNumberOfMoves = 0;
return list;
}
static void visitedInitilise(boolean visited[]) {
for (int i = 0; i <= (visited.length - 1); i++)
visited[i] = false;
}
public static void main(String args[]){
String testCaseID = "15"; // Write a test file number from 1 to 15, or
// ALL
TestCases.test(testCaseID);
}
}
歡迎的StackOverflow!你似乎沒有代碼的算法問題。如果您有解決方案,請將其包含在您的問題中。 –
對不起,「只是提供算法」仍然轉化爲:「請爲我作功課的核心部分」。這不是SO的工作原理。你應該自己處理這個問題並向我們展示。我們**幫助您解決您的問題;但大多數時候,沒有人願意爲你做**作業。 – GhostCat