LC260. 只出现一次的数字 III
//做了很多道只出现多少次的数字，都采用哈希映射采用空间换时间

class Solution {
    public int[] singleNumber(int[] nums) {
        Map<Integer, Integer> freq = new HashMap<Integer, Integer>();
        for (int num : nums) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }
        int[] ans = new int[2];
        int index = 0;
        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
            if (entry.getValue() == 1) {
                ans[index++] = entry.getKey();
            }
        }
        return ans;
    }
}

LC117. 填充每个节点的下一个右侧节点指针 II
//树的层次遍历采用队列。

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if(root == null)
        return null;
        Queue<Node> queue = new LinkedList();
        queue.offer(root);
        while(!queue.isEmpty()){
            int len = queue.size();
            Node pre = null;
            while(len != 0){
                Node node = queue.poll();
                if(pre != null){
                    pre.next = node;
                }
                pre = node;
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
                len--;
            }
        }
        return root;
    }
}

LC395. 至少有 K 个重复字符的最长子串
//要求这个最长子串的每个字符的重复长度都要大于k，那么我们可以考虑分治算法，将大的s字符串进行分割，我们将每个字符小于2的子字符串进行拆分，从不包含该字符的子字符串中进行查找。

class Solution {
    public int longestSubstring(String s, int k) {
        if(s.length() < k)
        return 0;
        HashMap<Character,Integer> counter = new HashMap();
        for(char ch : s.toCharArray()){
            counter.put(ch,counter.getOrDefault(ch,0) + 1);
        }
        for(char ch : counter.keySet()){
            if(counter.get(ch) < k){
                int res = 0;
                for(String split : s.split(String.valueOf(ch))){
                    res = Math.max(res,longestSubstring(split,k));
                }
                return res;
            }
        }
        return s.length();
    }
}