以下这个解法也是参考了一些讨论：
https://leetcode.com/discuss/110235/c-solution-using-euclidean-algorithm
还有这个解释原理的，没有看懂：https://leetcode.com/discuss/110525/a-little-explanation-on-gcd-method
class Solution { int gcd(int a, int b) { // below suppose a <= b // and if b < a, then b%a == b return a==0?b:gcd(b%a, a); } public: bool canMeasureWater(int x, int y, int z) { // notice, operator priority: % > == > && > || // so below is equivalent to // (z == 0) || ((z <= (x+y)) && ((z % gcd(x,y)) == 0)) return z == 0 || z <= (x+y) && z % gcd(x, y) == 0; } };
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31 / 31 test cases passed.
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Status:
Accepted |
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Runtime: 0 ms
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原文：http://www.cnblogs.com/charlesblc/p/5620143.html