https://leetcode.com/problems/water-and-jug-problem/description/ -- 365
There are two methods to solve this problem : GCD(+ elementary number theory) --> how to get GCF, HCD, BFS
Currently, I sove this by first method
1. how to compute GCD recursively
//get the GCD of two number s
int GCD(int a, int b){
if(a == 0) return b;
if(b == 0) return a;
return GCD(b,a%b);
}
12, 8 -> 8,4 -> 4, 4 -> 4, 0
math solution
Bézout‘s identity (also called Bézout‘s lemma) is a theorem in the elementary theory of numbers:
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
In addition, the greatest common divisor d is the smallest positive integer that can be written as ax + by
every integer of the form ax + by is a multiple of the greatest common divisor d.
If a or b is negative this means we are emptying a jug of x or y gallons respectively.
Similarly if a or b is positive this means we are filling a jug of x or y gallons respectively.
x = 4, y = 6, z = 8.
GCD(4, 6) = 2
8 is multiple of 2
so this input is valid and we have:
-1 * 4 + 6 * 2 = 8
In this case, there is a solution obtained by filling the 6 gallon jug twice and emptying the 4 gallon jug once. (Solution. Fill the 6 gallon jug and empty 4 gallons to the 4 gallon jug. Empty the 4 gallon jug. Now empty the remaining two gallons from the 6 gallon jug to the 4 gallon jug. Next refill the 6 gallon jug. This gives 8 gallons in the end)
code:
class Solution {
public boolean canMeasureWater(int x, int y, int z) {
//check the limitiation which x + y < z such as 3,4 , 8: notmeeting the requirement
if(x+ y < z) return false;
//check all 0
System.out.println(GCD(x,y));
//there is a theory about that
//ax + by = gcd z%gcd == 0 Bézout‘s identity
if(GCD(x,y) == 0) return z==0;
else return (z%GCD(x,y)==0);
}
//get the GCD of two number s
int GCD(int a, int b){
if(a == 0) return b;
if(b == 0) return a;
return GCD(b,a%b);
}
}
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BFS method
原文地址：https://www.cnblogs.com/stiles/p/leetcode365.html