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1002.查找常用字符


思路导图 先用几个二维数组保存 abcd 用ASCII码保存在二维数组中

class Solution {
     public List<String> commonChars(String[] A) 
	{
		List<String> list = new ArrayList<String>();
		int[][] arr = new int[A.length][26];
		boolean bol = true ;
		//将所有数据用二维数组保存
		for (int x = 0; x < A.length ; x++) 
		{
			for (int i = 0; i < A[x].length(); i++) 
			{
        //		System.out.println((A[x].charAt(i) - 'a')+" <-" + A[x].charAt(i));
				arr[x][A[x].charAt(i) - 'a']++;
			}
		}
//开始逐个元素进行查看
		for (int x = 0; x < 26; x++) 
		{	
			int count = arr[A.length - 1][x];
			if (arr[A.length - 1][x] != 0) 
			{
				for (int i = 0; i < arr.length - 1; i++) 
				{
					if (arr[i][x] == 0)  //如果有一个字符串某元素个数为0 不满足不成立
					{	
						bol = false;
						break;
					}
					count = Math.min(arr[i][x],count);//统计每个字符在每个字符串中出现次数
				}
			}
			if(bol ) 
			{
				for (int temp = 0; temp < count; temp++) 
				{	
				//	System.out.print((char)(x + 97));
					list.add(""+(char)(x + 97));
				}	
				
			}
			count=0;
			bol = true;
		}
		System.out.println();
		for(int i = 0;i<arr.length;i++) {
			for(int x = 0;x<26;x++) {
				//System.out.print((char)(x+97)+"->"+arr[i][x]+" ");
			}
//System.out.println();
		}
		return list;
	}
}

HashMap 一开始写不出来 卡了很久

class Solution {
    public List<String> commonChars(String[] A) {
        List<String> result = new ArrayList();
        List<HashMap<Character, Integer>> list = new ArrayList();
        for (String word: A) {
            HashMap<Character, Integer> table = new HashMap();//以字符串的形式保存到HashMap中
            for (int i = 0; i < word.length(); i++) {
                char ch = word.charAt(i);
                table.put(ch, table.getOrDefault(ch, 0) + 1);//新的getOrDefault()方法提供一个快捷的方式获取Map中的值
            }
            list.add(table);
        }
        HashMap<Character, Integer> map = list.get(0);
        for (Character ch: map.keySet()) {
            boolean flag = true;
            int min = map.get(ch);
            for (int i = 1; i < list.size(); i++) {
                HashMap<Character, Integer> charCnt = list.get(i);
                if (!charCnt.containsKey(ch)) {
                    flag = false;
                    break;
                }
                if (charCnt.get(ch) < min) {
                    min = charCnt.get(ch);
                }
            }
            if (flag) {
                for (int j = 0; j < min; j++) {
                    result.add("" + ch);
                }
            }
        }
        return result;
    }
}
posted @ 2019-07-08 12:52  cznczai  阅读(174)  评论(0编辑  收藏  举报