[英]Binary search of sorted-by-column 2D array only searching first column
因此,我对编程还很陌生,我正在尝试将二进制搜索算法应用于按列排序的二维数组。 我的程序仅在第一列上正确执行,因此我认为它陷入了无限循环。 问题似乎在while循环的第二个else-if语句之内,但是我对问题可能是完全不知所措。
public static void main(String[] args) {
// TODO Auto-generated method stub
int[][] array = { {5, 8, 10, 6},
{20, 10, 20, 15},
{20, 15, 25, 20},
{20, 20, 32, 25} };
int query = 20;
int search_status;
search_status = count(array, query);
System.out.println(query + " occurred " + search_status + " times");
}
public static int count(int[][]array, int query){
int count = 0;
int low = 0;
int high = array.length - 1;
for (int c = 0; c < array[0].length; c++) {
while (low <= high) {
int mid = low + (high - low) / 2;
if (array[mid][c] > query) {
high = mid - 1;
}
else if (array[mid][c] < query) {
low = mid + 1;
}
else if (array[mid][c] == query) {
count++;
int up = -1;
int down = 1;
while ((mid + up >= 0) && (array[mid + up][c] == query)) {
up--;
count++;
}
while ((mid + down <= array.length - 1) && (array[mid + down][c] == query)) {
down++;
count++;
}
return count;
}
}
}
return - 1;
}
}
尝试以下将列首先推入行的方法:
public static void main(String[] args) {
// TODO Auto-generated method stub
int[][] array = { {5, 8, 10, 6},
{20, 10, 20, 15},
{20, 15, 25, 20},
{20, 20, 32, 25} };
int query = 20;
int search_status = 0;
for (int c = 0; c < array.length; c++)
search_status += count(array, query, c);
System.out.println(query + " occurred " + search_status + " times");
}
public static int count(int[][]array, int query, int row){
int[] column = new int[array.length];
int count = 0;
int low = 0;
int high = column.length - 1;
for (int i = 0; i < array[row].length; i++)
column[i] = array[i][row];
while (low <= high) {
int mid = low + (high - low) / 2;
if (column[mid] > query) {
high = mid - 1;
}
else if (column[mid] < query) {
low = mid + 1;
}
else if (column[mid] == query) {
int mover = -1;
int counted = 1;
// Go all the way down
while ((mid + mover >= 0) && (column[mid + mover] == query))
{
mover--;
counted++;
}
mover+=counted+1;
count+=counted;
while ((mid + mover <= column.length - 1) && (column[mid + mover] == query))
{
mover++;
count++;
}
return count;
}
}
return 0;
}
编辑 :稍微更好的实现,以免重复计算。
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