780 Reaching Points
780. Reaching Points
1. Question
A move consists of taking a point(x, y)
and transforming it to either(x, x+y)
or(x+y, y)
.
Given a starting point(sx, sy)
and a target point(tx, ty)
, returnTrue
if and only if a sequence of moves exists to transform the point(sx, sy)
to(tx, ty)
. Otherwise, returnFalse
.
Examples:
Input: sx = 1, sy = 1, tx = 3, ty = 5
Output: True
Explanation:
One series of moves that transforms the starting point to the target is:
(1, 1) -> (1, 2)
(1, 2) -> (3, 2)
(3, 2) -> (3, 5)
Input: sx = 1, sy = 1, tx = 2, ty = 2
Output: False
Input: sx = 1, sy = 1, tx = 1, ty = 1
Output: True
Note:
sx, sy, tx, ty
will all be integers in the range[1, 10^9]
.
2. Implementation
(1) BFS (TLE)
class Solution {
public boolean reachingPoints(int sx, int sy, int tx, int ty) {
if (sx > tx || sy > ty) {
return false;
}
Queue<int[]> queue = new LinkedList<>();
queue.add(new int[] {sx, sy});
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] curPoint = queue.remove();
int x = curPoint[0], y = curPoint[1];
if (x == tx && y == ty) {
return true;
}
if (x + y <= tx) {
queue.add(new int[] {x + y, y});
}
if (x + y <= ty) {
queue.add(new int[] {x, x + y});
}
}
}
return false;
}
}
(2) Math
思路:看解析
class Solution {
public boolean reachingPoints(int sx, int sy, int tx, int ty) {
while (sx < tx && sy < ty) {
if (tx > ty) {
tx %= ty;
}
else {
ty %= tx;
}
}
if (sx == tx) {
return (ty - sy) % sx == 0;
}
else if (sy == ty) {
return (tx - sx) % sy == 0;
}
else {
return false;
}
}
}
3. Time & Space Complexity
Math: 时间复杂度O(max(tx, ty)), 空间复杂度O(1)
Last updated
Was this helpful?