Question

问题是：给定仅由小写字母组成的字符串数组A，返回列表中所有字符串（包括重复项）中显示的所有字符的列表。例如，如果某个字符在所有字符串中出现3次，但没有出现4次，那么您需要在最终答案中将该字符包含3次。

您可以按任何顺序返回答案。

示例1：

输入：[“ bella”，“ label”，“ roller”] 输出：[“ e”，“ l”，“ l”]

示例2：

输入：[“ cool”，“ lock”，“ cook”] 输出：[“ c”，“ o”]

注意：

1 <=长度<= 100

1 <= A [i]。长度<= 100

A [i] [j]是小写字母

这是我的代码：

class Solution {
    public List<String> commonChars(String[] A) {
        List<String> charArr = new ArrayList<>();

        for(int i = 1; i<A.length ; i++){
            A[0].replaceAll("[^" + A[i] + "]", "");
        }
        for(int i=0; i<A[0].length(); i++){
            charArr.add(Character.toString(A[0].charAt(i)));
        }
        return charArr;

    }
}

我得到的结果

输入： [“ bella”，“ label”，“ roller”]

输出： [“ b”，“ e”，“ l”，“ l”，“ a”]

预期： [“ e”，“ l”，“ l”]

显然字符没有被删除，有人可以帮助我解决这个问题吗？

Answer 1

以下是我如何解决此问题的伪代码：

Place the letters from the first string of the array into a list
for each letter in the list{
    count the occurrences of the letter
    for each string in the array{
        count the occurrences of the letter
        if the letter occurs in the list more often than in the string{
            decrease occurrence of letter in list to match letter in string
        }
    }
}

希望这会有所帮助！

Answer 2

不确定java，但这是python中的1个内衬

list(functools.reduce(lambda a,b: a&b, [Counter(c) for c in A]).elements())

>>> from collections import Counter
>>> import functools

>>> A = ["bella","label","roller"]
>>> list(functools.reduce(lambda a,b: a&b, [Counter(c) for c in A]).elements())
['e', 'l', 'l']
>>> A =  ["cool","lock","cook"]
>>> list(functools.reduce(lambda a,b: a&b, [Counter(c) for c in A]).elements())
['c', 'o']
>>>

Answer 3

我用Manaar的想法解决了这个问题，谢谢大家的建议和评论！

$(document).ready(function(){
    var media = $('video');

    function checkmedia(){
        // Get current browser top and bottom
        var scrollTop = $(window).scrollTop();
        var scrollBottom = $(window).scrollTop() + $(window).height();

        media.each(function(index, el) {
           var height = parseInt($(this).height());
           var visiblechunk = height/1.2;
           var vidpostop = $(this).offset().top;
           var vidposbottom = $(this).height() + vidpostop;

            /* Only play when a certain portion of the video is visible */
            vidpostop = vidpostop+visiblechunk;
            vidposbottom = vidposbottom - visiblechunk;
            if(scrollTop < vidposbottom && scrollBottom > vidpostop){
                if($(this).hasClass('autoplayed')){

                }else{
                   $(this).get(0).play();
                   $(this).addClass('autoplayed');
                }

            }else{
                $(this).get(0).pause();
                $(this).removeClass('autoplayed');
            }
        });
    }
    $(window).scroll(function(){
       checkmedia();
    })
})

Answer 4

略有改动的版本：

public List<String> commonChars01(String[] A) {
    int[][] lowerCase = new int[26][A.length];
    
    for(int i = 0 ; i< A.length ; i++){
        for(int j = 0; j < A[i].length(); j++) {
            lowerCase[A[i].charAt(j) -'a'][i] ++;
        }                        
    }        
  
    List<String> list = new ArrayList<>();
    
    for (int i = 0; i < 26; i++) { 
        // all the elements of the array should be equal
        long count = IntStream.of(lowerCase[i])
                .boxed().distinct().count();
        
        // [0, 0, 0] should fail, hence sum should be greater than or equal to 3.
        int sum = IntStream.of(lowerCase[i])
                .boxed().reduce(0, (a,b) -> a + b); 
                
        if ( sum >= 3 && count == 1) {
            int timesRepeated = lowerCase[i][0];
            while (timesRepeated != 0) {
                list.add(Character.toString((char)(i + 'a')));
                timesRepeated--;
            }
        }
    }        
    
    return list;
}

Leetcode1002。查找字符串之间的通用字符-Java

4 个答案: