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)
, return True
if and only if a sequence of moves exists to transform the point (sx, sy)
to (tx, ty)
. Otherwise, return False
.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]
.
Solution #1, 因为给定的都是正数,所有可以对比tx跟ty,大的减去小的就是之前的那一对数
class Solution { public boolean reachingPoints(int sx, int sy, int tx, int ty) { while (tx >= sx && ty >= sy) { if (tx == sx && ty == sy) return true; if (tx > ty) { tx -= ty; }else { ty -= tx; } } return false; } }
Solution #2, 跟#1类似,但是我们需要快速收敛。考虑以下例子
2, 12
2, 8
2, 6
2, 4
2, 2
class Solution { public boolean reachingPoints(int sx, int sy, int tx, int ty) { while (tx >= sx && ty >= sy) { if (tx == sx && ty == sy) return true; if (tx > ty) { if (ty == sy) return (tx - sx) % ty == 0; tx %= ty; }else { if (tx == sx) return (ty - sy) % tx == 0; ty %= tx; } } return false; } }
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