发表于

1 自定义排序

class Solution {
    public String largestNumber(int[] nums) {
        int n=nums.length;
        Integer[] arr=new Integer[n];
        for(int i=0;i<n;i++){
            arr[i]=nums[i];
        }
        //从大到小排序
        Arrays.sort(arr,(x,y)->{
            long sx=10,sy=10;
            while(sx<=x){
                sx=sx*10;
            }
            while(sy<=y){sy=sy*10;}

            return (int) (y*sx+x-(x*sy+y));
        });

        if(arr[0]==0){
            return "0";
        }
        StringBuilder sb = new StringBuilder();
        for(int num:arr){
            sb.append(num);
        }
        return sb.toString();
    }
}

2 数组

//二维数组
List<int[]>[] g=new ArrayList[n];
Arrays.setAll(g,i->new ArrayList<>());


//排序
int[][] points;
Arrays.sort(points,(a,b)->Integer.compare(a[0],b[0]));

int[][] grid;
int m=grid.length,n=grid[0].length;
int[] row=new int[m];
int[] col=new int[n];


int[] nums=new int[n];
// Arrays.stream(nums).sum();
int n=nums.length;
Arrays.fill(nums,1);


return new int[]{-1,-1};


//数组排序
int[] nums;
Arrays.sort(nums);
操作 说明
nums.set(index,last) 设置Index下标元素值
nums.remove(index)
nums.size()

3 List

//二维数组
List<List<Integer>> ans=new ArrayList<List<Integer>>();

4 String

String s;
//获取长度
s.length();

//某一下标字符
s.charAt(index);

//去掉最后一个字符
StringBuilder s=new StringBuilder();
s.deleteCharAt(s.length()-1);

5 Map

操作 说明
s.containsKey(val)
s.remove(key)
s.get(key) 
Map<Character, Character> pairs = new HashMap<Character, Character>() {{
          put(')', '(');
          put(']', '[');
          put('}', '{');
      }};

6 Set

操作 说明
Set<Character> s=new HashSet()
s.contains(ch)
s.add(ch) s中包含ch，返回false
s.remove(ch)

7 Stack

 Deque<Character> stack=new LinkedList<Character>();
 stack.isEmpty()
 stack.peek()
 stack.pop()
 stack.push(ch)