字符流中第一个不重复的字符

1.题目

2.解法1：每次都遍历：

public class Solution {
    // String str不赋初始值为null，str += ch, ch = 'h', 会变成nullh
    private String str = "";
    private int[] arr = new int[256];

    //Insert one char from stringstream
    public void Insert(char ch) {
        str += ch;
        arr[ch]++;
    }

    //return the first appearence once char in current stringstream
    public char FirstAppearingOnce() {
        for (int i = 0; i < str.length(); i++) {
            if (arr[str.charAt(i)] == 1) {
                return str.charAt(i);
            }
        }
        return '#';
    }
}

2.解法2（减少遍历次数，队列，先入先出）

字符流是动态变化的，每次都是往后插入一个元素，需要返回第一个出现1次的元素，那么需要队列来处理

import java.util.ArrayDeque;

public class Solution {
    private ArrayDeque<Character> fArr = new ArrayDeque<>();
    private int[] count = new int[256];

    //Insert one char from stringstream
    public void Insert(char ch) {
        // 如果第一次出现就添加进去
        if(count[ch]++ == 0) {
            fArr.offer(ch);
        }
    }

    //return the first appearence once char in current stringstream
    public char FirstAppearingOnce() {
        Character c;
       while((c = fArr.peek()) != null) {
            // 判断此字符的count是不是为1
           if(count[c.charValue()] == 1) {
               return c;
           }
           // 如果不为1，移除
           fArr.remove();
       }
        return '#';
    }
}

时间复杂度为O(n),空间复杂度为O(n)