[leedcode][JAVA][365][BFS]

【Problem Description】

有两个容量分别为 x升 和 y升 的水壶以及无限多的水。请判断能否通过使用这两个水壶，从而可以得到恰好 z升 的水？

如果可以，最后请用以上水壶中的一或两个来盛放取得的 z升 水。

你允许：

装满任意一个水壶
清空任意一个水壶
从一个水壶向另外一个水壶倒水，直到装满或者倒空
示例 1: (From the famous "Die Hard" example)

输入: x = 3, y = 5, z = 4
输出: True
示例 2:

输入: x = 2, y = 6, z = 5
输出: False

[Thinking] answer

1. BFS breadth-first traversal

• set to ensure that the situation does not repeat
• queue through all the cases

public boolean canMeasureWater_BFS(int x, int y, int z) {
    if (z < 0 || z > x + y) {
        return false;
    }
    Set<Integer> set = new HashSet<>();
    Queue<Integer> q = new LinkedList<>();
//巧妙过度
    q.offer(0);
//遍历所有情况 第一次 x,y入队列
    while (!q.isEmpty()) {
        int n = q.poll();
//巧妙限制条件！
        if (n + x <= x + y && set.add(n + x)) {
            q.offer(n + x);
        }
        if (n + y <= x + y && set.add(n + y)) {
            q.offer(n + y);
        }
        if (n - x >= 0 && set.add(n - x)) {
            q.offer(n - x);
        }
        if (n - y >= 0 && set.add(n - y)) {
            q.offer(n - y);
        }
        if (set.contains(z)) {
            return true;
        }
    }
    return false;
}

2. Mathematical Methods greatest common divisor

public boolean canMeasureWater(int x, int y, int z) {
    if (z == 0) return true;
    if (x + y < z) return false;

    int big = Math.max(x, y);
    int small = x + y - big;

    if (small == 0) return big == z;

//最大公约数 
    while (big % small != 0) {
        int temp = small;
        small = big % small;
        big = temp;
    }
    return z % small == 0;
}

【to sum up】

1. want to clear all possible, to achieve dot columns, use containers, clever use of border

2. Queue operation

• Add to add one yuan cable if the queue is full, throw an exception IIIegaISlabEepeplian
• Remove to remove the head of the queue and returns the element if the queue is empty, an exception is thrown a NoSuchElementException
• element head of the queue element is returned if the queue is empty, an exception is thrown a NoSuchElementException
• offer and add an element returns true if the queue is full, false is returned
• Remove and return an element poll asked the head of the queue if the queue is empty or null
• Returns peek head of the queue If the queue element is empty, or null
• If you add an element to put the queue is full, then blocked
• take removes and returns the element head of the queue

3.Collections

Set and List have inherited Conllection, Map no
methods interface Collection :

• boolean add (Object o): An object is added to the reference in the set
• void clear (): Delete all the objects in the collection that no longer holds these objects references
• boolean isEmpty (): determines whether the set is empty
• boolean contains (Object o): determining whether a specific object in the collection holding references
• Iterartor iterator (): Returns an Iterator objects, can be used to traverse the elements in the collection
• boolean remove (Object o): Removes an object from a set of reference
• Returns the number of elements in the set: int size ()
• Object [] toArray (): returns an array, the array comprises all of the elements in the collection
• About: Iterator () and toArray () methods are used for all elements of the set, the former returns an Iterator objects, which returns a collection of all elements in the array contains.
  Iterator interface declares the following methods :
• hasNext (): determining whether a set of elements traversed completed, if not, returns true
• next (): returns the next element
• remove (): delete from the collection a next () method returns the element.
  Set the add () method is how to determine whether the object has been stored in the collection?

    boolean isExists=false;
    Iterator iterator=set.iterator();
    while(it.hasNext())           {
    String oldStr=it.next();
    if(newStr.equals(oldStr)){
    isExists=true;
    }