You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.
If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.
Operations allowed:
- Fill any of the jugs completely with water.
- Empty any of the jugs.
- Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.
**Example 1: **
Input: x = 3, y = 5, z = 4
Output: True
Example 2:
Input: x = 2, y = 6, z = 5
Output: False
这道题实际上是道数学智力题，求解的关键就是判断z能否整除x与y的最大公约数。当然，这道题有些隐藏约束，就是z<=x+y以及z可以为0，当然多过几遍测试用例基本上就能察觉了。上代码：
/**
* @param {number} x
* @param {number} y
* @param {number} z
* @return {boolean}
*/
var canMeasureWater = function (x, y, z) {
const findDivisor = function (a, b) {
while (b) {
let c = a % b;
a = b;
b = c;
}
return a;
};
const divisor = findDivisor(x, y);
if (z > x + y) {
return false;
}
else if (z === x || z === y || z === 0) {
return true;
}
return (z % divisor === 0);
};
需要一些优化的地方可能就是找最大公约数了，这里我选用了辗转相除的方法。当然也有辗转相减和暴力循环的方式。具体算法可以自行百度一下。
