Difference between revisions of "TADM2E 4.35"
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Having M[0..n-1][0..m-1] and a struct Point {int x, int y}, we could have the following solution: | Having M[0..n-1][0..m-1] and a struct Point {int x, int y}, we could have the following solution: | ||
| − | Point* findPosition(int key) { | + | |
| − | + | Point* findPosition(int key) { | |
| − | + | int row = 0, col = m-1; | |
| − | + | while (row < n && col >= 0) { | |
| − | + | if (M[row][col] == key) { | |
| − | + | return new Point(row,col); | |
| − | + | } | |
| − | + | else if (M[row][col] > key) { | |
| − | + | col--; | |
| − | + | } | |
| − | + | else row++; | |
| − | + | } | |
| − | } | + | return NULL; |
| + | } | ||
Latest revision as of 13:00, 30 April 2015
$ O(n+m) $ is necessary and sufficient. Lower bound comes from potentially independent values along second diagonal -- upper bound comes from observing that we can eliminate either a row or a column in each comparison if we start from the lower left corner and walk up or left.
Having M[0..n-1][0..m-1] and a struct Point {int x, int y}, we could have the following solution:
Point* findPosition(int key) {
int row = 0, col = m-1;
while (row < n && col >= 0) {
if (M[row][col] == key) {
return new Point(row,col);
}
else if (M[row][col] > key) {
col--;
}
else row++;
}
return NULL;
}
