Java各种初始化

  • Z时代
  • 2024-01-10
  • 分类：综合

java

  • int[] res = new int[]{-1, -1};

  • String res = “”;

  • 输出类型List<List>如何被初始化？List<List> res = new ArrayList<>(); 有关这些复杂类型的初始化还有待于进一步研究

  • List res = new ArrayList<>();//不确定长度的list

  • 在Java中 是不是List用[] 但是Array用{}

  • 二维数组： boolean[][] dp = new boolean[s.length()][s.length()];//二维数组也是array,因此初始化的时候要么长度给定 要么直接初始化元素，见下：

  • 二维数组：int[][] a = {{1,2}{3,4}}

  • HashMap<Integer, Integer> map = new HashMap<>();

  • 值得注意的是 hashmap不能直接用值进行初始化（？）只能用map.put(k,v)进行放入值

  • HashSet set = new HashSet<>();

  • array的几种初始化方式：int[] a; int b[]; //数组的声明 int[] array1 = {4,5,6}; int[] array2 = new int[3]; int[] array3 = new int[]{1,2,3}//数组的初始化

  • 输出链表初始化：ListNode dummy = new ListNode(0); 而且要注意的是 链表的使用一般都是要链表指针的 如下：ListNode cur = dummy;//链表的遍历是通过指针来完成

  • StringBuilder类型数据初始化： StringBuilder res = new StringBuilder();

  • 数组型StringBuilder类型数据初始化：StringBuilder[] res = new StringBuilder[number_of_rows];//与数组结合

  • Stack初始化Stack<Character> stack = new Stack<Character>();

  • PQ初始化PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (a, b) -> a.val - b.val);或者Queue<TreeNode> queue = new LinkedList<>(); OR: PriorityQueue<Integer> heap = new PriorityQueue<Integer>((n1, n2) -> map.get(n1) - map.get(n2));//and heap will be map with sorted in its value

  • Deque的初始化：Deque<Integer> deque = new LinkedList<>();

  • 二维数据的初始化总结如下：(理论上来说 能用Array的都可以用ArrayList去做 只是有时候输出的要求已经定好 最后必须进行相关转化)

  • 内层外层都需要ArrayList(即内外层长度都不定): 这种情况很常见，直接List<List<Integer>> res = new ArrayList<>(); List<Integer> list = new ArrayList<>();

  • 内层外层都需要Array（内层外层长度都固定）: int[][] res = new int[length1][length2]; int[] list = new int[length];

  • 外层ArrayList,内层Array(外层长度不固定 内层长度固定 比如LC56): List<int[]> list = new ArrayList<>();

  • 外层Array,内层ArrayList(外层长度固定 内层长度不固定，比如LC347 Top K frequency elements): List<Integer>[] array = new List[length];//it’s like int[] but instead of int,we have List

  • 此外 如何把这些二维数据的格式之间相互转化？Arrays.asList(), var.toArray()看起来没什么用处

  • random object: ‘Random rand = new Random();’

  • sort String array based on it length(small->larger):

  • Arrays.sort(words, new Comparator<String>() {
    @Override
    public int compare(String s1, String s2) {return s1.length() - s2.length();}
    });

  • sort 2d array based on the first column:

  • Arrays.sort(words, new Comparator<int[]>() {
    @Override
    public int compare(int[] interval1, int[] interval2) {return interval[0] - interval[1];}
    }); //not sure if this gonna work

  • another way:

  • Arrays.sort(intervals, (a, b) -> (a[0]- b[0]));

  • Min heap and max Heap:PriorityQueue<Integer> minheap = new PriorityQueue<>()PriorityQueue<Integer> maxheap = new PriorityQueue<>(Comparator.reverseOrder()); or maxheap can be initialize as :

  • PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>( new Comparator<Integer>() { public int compare(Integer i1, Integer i2) { return i2.compareTo(i1); } } );

  • convert hashmap key set to list: List<String> res_files = new ArrayList<>(t.files.keySet())

  • we want to sort a list of Node, and nodes have attribite of times and sentence. we want all those node sort in times first, if they are equal, then we sort them in sentence: Collections.sort(list, (a, b) -> a.times == b.times? a.sentence.compareTo(b.sentence): b.times - a.times);

  • initialize a hashset with values: List<Character> list = Arrays.asList(\'a\',\'e\',\'i\',\'o\',\'u\'); HashSet<Character> set = new HashSet<>(list); or you can use Collections.addAll(), or you can initialize a set and add items one by one.

  • String to character array: s.toCharArray(), charArray to String: new String(char[] chars)

  • how to traverse hashset? just use for each, or use iterator.

以上是 Java各种初始化 的全部内容， 来源链接： utcz.com/z/392752.html

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