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365. Water and Jug Problem
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: (From the famous "Die Hard" example)
Input: x = 3, y = 5, z = 4 Output: True
Example 2:
Input: x = 2, y = 6, z = 5 Output: False
Credits:
Special thanks to @vinod23 for adding this problem and creating all test cases.
/**
* Math problem
* 能测的体积是两个瓶子最大公约数的倍数
* 比较麻烦的是0 的情况!!
*/
/** * Math problem * 能测的体积是两个瓶子最大公约数的倍数 * 比较麻烦的是0 的情况!! * 可用数论的知识 Bézout‘s identity 求最大公约数 */ public class Solution { public boolean canMeasureWater(int x, int y, int z) { if(x+y < z) return false; if(x == 0 && y == 0 && z == 0) return true; if(x == 0 || y == 0 && z != 0) return false; return z % gcd1(x,y) == 0; } // Bézout‘s identity //let a and b be nonzero integers and let d be their greatest common divisor. Then there exist integers x and y such that ax+by=d public int gcd1(int x, int y){ while(y != 0){ int temp = y; y = x % y; x = temp; } return x; } public int gcd2(int x, int y){ int res = 1; int end = Math.min(x, y); if(end == 1) return 1; for(int i = 2; i <= end; i++){ if(x % i == 0 && y % i == 0) res = i; } return res; } }
365. Water and Jug Problem
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