LeetCode 1170.比较字符串最小字母出现频次

LeetCode 1170.比较字符串最小字母出现频次

题目

题解

写一个方法，返回字符串的最小字符出现次数。

代码

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] answer=new int[queries.length];
        
        for(int i=0;i<queries.length;i++){
            answer[i]=0;
            for(int j=0;j<words.length;j++){
                if(function(queries[i])<function(words[j])){
                    answer[i]++;
                }
            }
        }
        return answer;
    }
    public int function(String str)
    {
        char minChar=str.charAt(0);
        int minCount=1;
        for(int i=1;i<str.length();i++)
        {
          if(str.charAt(i)<minChar)
          {
            minChar=str.charAt(i);
            minCount=1;
          }
            else if(str.charAt(i)==minChar){
            minCount++;
            }
            else {
                
            }
        }
        return minCount;
    }
}

题解二

优化了一下，把字符串数组的结果在循环外用数组储存，不必在循环内一遍遍调用函数。

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] answer=new int[queries.length];
        int[] dataq=new int[queries.length];
        for(int i=0;i<queries.length;i++){
            dataq[i]=function(queries[i]);
        }
        int[] dataw=new int[words.length];
        for(int i=0;i<words.length;i++){
            dataw[i]=function(words[i]);
        }
        for(int i=0;i<queries.length;i++){
            answer[i]=0;
            for(int j=0;j<words.length;j++){
               if(dataq[i]<dataw[j]){
                    answer[i]++; 
               }
            }
        }
        return answer;
    }
    public int function(String str)
    {
        char minChar=str.charAt(0);
        int minCount=1;
        for(int i=1;i<str.length();i++)
        {
          if(str.charAt(i)<minChar)
          {
            minChar=str.charAt(i);
            minCount=1;
          }
            else if(str.charAt(i)==minChar){
            minCount++;
            }
            else {
                
            }
        }
        return minCount;
    }
}