如何以词法方式订购子串

Question

我想按字母顺序排列字符串“s”的子串，长度为“k”

我尝试首先使用comapareTo函数对字符串的字符进行词法排序，然后尝试打印第一个和最后一个子字符串

public static String getSmallestAndLargest(String s, int k) {
    String smallest = "";
    String largest = "";
    char ch1,ch2,temp;
    int i,j,res; 

    // 'smallest' must be the lexicographically smallest substring of length 'k'
    // 'largest' must be the lexicographically largest substring of length 'k'
    for(i=0;i<s.length();i++)
    {
        ch1=s.charAt(i);
        for(j=i+1;j<=s.length();j++)
        {
            ch2=s.charAt(j);
            res=ch2.compareTo(ch1);
            if(res<0)
            {
                temp=ch2;
                ch2=ch1;
                ch1=temp;
            }
        }
    }
    smallest=s.substring(0,k);
    largest=s.substring(s.length()-k);
    return smallest + "\n" + largest;
}

预期输出：将字典上最小和最大的子字符串作为单个换行符分隔的字符串返回。

input: welcometojava
3

expected output:ava
wel

Answer 1

我们初始化max和min作为大小为k的第一个子字符串。我们通过删除前一个子字符串的第一个字符并添加新字符串的最后一个字符来遍历剩余的子字符串。我们跟踪字典上最大和最小的。

public class GFG { 

public static void getSmallestAndLargest(String s, int k) 
{ 
    // Initialize min and max as first substring of size k 
    String currStr = s.substring(0, k); 
    String lexMin = currStr; 
    String lexMax = currStr; 

    // Consider all remaining substrings. We consider 
    // every substring ending with index i. 
    for (int i = k; i < s.length(); i++) { 
        currStr = currStr.substring(1, k) + s.charAt(i); 
        if (lexMax.compareTo(currStr) < 0)      
             lexMax = currStr; 
        if (lexMin.compareTo(currStr) > 0) 
             lexMin = currStr;             
    } 

    // Print result. 
    System.out.println(lexMin); 
    System.out.println(lexMax); 
} 

// Driver Code 
public static void main(String[] args) 
{ 
    String str = "GeeksForGeeks"; 
    int k = 3; 
    getSmallestAndLargest(str, k); 
} 

}

Answer 2

你的想法是对的，但你试图比较单个字符。相反，在每次迭代中，你应该取一个长度为k的子字符串，并将其与当前的“最小”和“最大”字符串进行比较：

public static String getSmallestAndLargest(String s, int k) {
    String curr = s.substring(0, k);
    String smallest = curr;
    String largest = curr;
    for (int i = 1; i < s.length() - k + 1; ++i) {
        curr = s.substring(i, i + k);
        if (smallest.compareTo(curr) > 0) {
            smallest = curr;
        }
        if (largest.compareTo(curr) < 0) {
            largest = curr;
        }
    }
    return smallest + "\n" + largest;
}