780. Reaching Points

最新推荐文章于 2021-12-24 19:38:27 发布

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最新推荐文章于 2021-12-24 19:38:27 发布

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文章标签： java

原文链接：http://www.cnblogs.com/ruruozhenhao/p/10872636.html

本文探讨了一种独特的点转换算法，该算法判断是否能通过特定的数学变换从初始点(sx, sy)到达目标点(tx, ty)。文章提供了详细的算法实现，并通过实例展示了算法的应用过程。

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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

Approach #1: Math. [Java]

class Solution {
    public boolean reachingPoints(int sx, int sy, int tx, int ty) {
        while (sx < tx && sy < ty) {
            if (ty > tx) ty %= tx;
            else tx %= ty;
        }
        return sx == tx && sy <= ty && (ty - sy) % sx == 0 ||
               sy == ty && sx <= tx && (tx - sx) % sy == 0;
    }
}

Analysis:

If we start from sx, sy, it will be hard to find tx, ty.

If we start from tx, ty, we can find only one path to go back to sx, sy.

First, if 2 target points are still bigger than 2 starting point, we reduce target points.

Second, check if reduce target points to (x, y + kx) or (x + ky, y).