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@KodeSeeker
Created January 29, 2018 06:16
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SpiralMatrix1
public List<Integer> spiralOrder(List<ArrayList<Integer>> a){
if( a == null || a.isEmpty()) {
return null;
}
// base case 3x3 then tweak
int startRow = 0;
int startCol = 0;
int endRow= a.size();
int endCol = a.get(0).size();
// terminating condition .
// if startRow == endRow and startCol == end Col then we print
//element a[startRow][Endcol], this is the element in the middle of the array
// if only one of the conditions meet then exit because we've covered the whole matrix.
while (startRow <=endRow && startCol <= endCol) {
//deal with matrices that have an element in the middle/ just one element left.
if (startRow == endRow) && (startCol == endCol) {
result.add(a.get(startRow).get(startCol));
break;
}
for (int i= startRow, int j = startCol; j<=endCol-1; j++) {
result.add(a.get(i).get(j));//1,2
}
for (int i= startRow, int j = endCol; i<=endRow-1; i++) {
result.add(a.get(i).get(j));//3,6
}
for (int i= endRow, int j = endCol; j>=startCol+1; j--) {
result.add(a.get(i).get(j));//9,8
}
for (int i= endRow, int j = startCol; i>=startRow+1; i--) {
result.add(a.get(i).get(j));//7,4
}
startRow++;
startCol++;
endRow --;
endCol --;
}
}
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