如何使用二进制搜索在排序数组中查找重复项？

Question

我试图通过重置高变量来扩展通过二进制搜索找到整数匹配数的函数，但是它陷入了循环。 我猜一种解决方法是复制此函数以获得最后一个索引来确定匹配数，但是我认为这不是一个很好的解决方案。

由此：

public static Matches findMatches(int[] values, int query) {
    int firstMatchIndex = -1;
    int lastMatchIndex = -1;
    int numberOfMatches = 0;

    int low = 0;
    int mid = 0;
    int high = values[values.length - 1];
    boolean searchFirst = false;

    while (low <= high){
        mid = (low + high)/2;

        if (values[mid] == query && firstMatchIndex == -1){
            firstMatchIndex = mid;

            if (searchFirst){
                high = mid - 1;
                searchFirst = false;
            } else { 
                low = mid + 1;
            }

        } else if (query < values[mid]){
            high = mid - 1;
        } else {
            low = mid + 1;
        }           
    }

    if (firstMatchIndex != -1) { // First match index is set
        return new Matches(firstMatchIndex, numberOfMatches);
    }
    else { // First match index is not set
        return new Matches(-1, 0); 
    }
}

对于这样的事情：

public static Matches findMatches(int[] values, int query) {
    int firstMatchIndex = -1;
    int lastMatchIndex = -1;
    int numberOfMatches = 0;

    int low = 0;
    int mid = 0;
    int high = values[values.length - 1];
    boolean searchFirst = false;

    while (low <= high){
        mid = (low + high)/2;

        if (values[mid] == query && firstMatchIndex == -1){
            firstMatchIndex = mid;

            if (searchFirst){
                high = values[values.length - 1]; // This is stuck in a loop
                searchFirst = false;
            } 
        } else if (values[mid] == query && lastMatchIndex == -1){
            lastMatchIndex = mid;

            if (!searchFirst){
                high = mid - 1;
            } else { 
                low = mid + 1;
            }
        } else if (query < values[mid]){
            high = mid - 1;
        } else {
            low = mid + 1;
        }

    }

    if (firstMatchIndex != -1) { // First match index is set
        return new Matches(firstMatchIndex, numberOfMatches);
    }
    else { // First match index is not set
        return new Matches(-1, 0); 
    }
}

Answer 1

您的代码有问题：

high = values[values.length - 1];

应该

high = values.length - 1;

另外，您不需要numberOfMatches和searchFirst之类的变量，我们可以有一个非常简单的解决方案。

现在来解决这个问题，我了解您想要什么，我认为二进制搜索适合这种查询。

达到要求的最佳方法是， 一旦找到匹配项，您就从该索引向前或向后前进，直到发生不匹配为止，这在计算firstMatchIndex和numberOfMatches方面既优雅又有效。

因此，您的功能应为：

public static Matches findMatches(int[] values, int query) 
{
 int firstMatchIndex = -1,lastMatchIndex=-1;
 int low = 0,mid = 0,high = values.length - 1;
 while (low <= high)
 {
      mid = (low + high)/2;

      if(values[mid]==query)
      {
          lastMatchIndex=mid;
          firstMatchIndex=mid;
          while(lastMatchIndex+1<values.length&&values[lastMatchIndex+1]==query)
           lastMatchIndex++;
          while(firstMatchIndex-1>=0&&values[firstMatchIndex-1]==query)
           firstMatchIndex--; 
          return new Matches(firstMatchIndex,lastMatchIndex-firstMatchIndex+1); 
      }
      else if(values[mid]>query)
       high=mid-1;
      else low=mid+1;
 }
 return new Matches(-1,0);
}          

Answer 2

您不能只使用set来查找重复项吗？

像这样：

package example;

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;

public class DuplicatesExample {

    public static void main(String[] args) {
        String[] strings = { "one", "two", "two", "three", "four", "five", "six", "six" };
        List<String> dups = getDups(strings);
        System.out.println("DUPLICATES:");
        for(String str : dups) {
            System.out.println("\t" + str);
        }
    }

    private static List<String> getDups(String[] strings) {
        ArrayList<String> rtn = new ArrayList<String>();
        HashSet<String> set = new HashSet<>();
        for (String str : strings) {
            boolean added = set.add(str);
            if (added == false ) {
                rtn.add(str);
            }
        }
        return rtn;
    }

}

输出：

DUPLICATES:
    two
    six

Answer 3

我将您的问题分为两部分-使用二进制搜索查找数字并计算匹配数。 第一部分由搜索功能解决，而第二部分由findMatches函数解决：

public static Matches findMatches(int[] values, int query) {

    int leftIndex = -1;
    int rightIndex = -1;
    int high = values.length - 1;

    int matchedIndex = search(values, 0, high, query);

    //if at least one match
    if (matchedIndex != -1) {

        //decrement upper bound of left array
        int leftHigh = matchedIndex - 1;
        //increment lower bound of right array
        int rightLow = matchedIndex + 1;

        //loop until no more duplicates in left array
        while (true) {

            int leftMatchedIndex = search(values, 0, leftHigh, query);

            //if duplicate found
            if (leftMatchedIndex != -1) {
                leftIndex = leftMatchedIndex;
                //decrement upper bound of left array
                leftHigh = leftMatchedIndex - 1;
            } else {
                break;
            }
        }

        //loop until no more duplicates in right array
        while(true){
            int rightMatchedIndex = search(values, rightLow, high, query);

            //if duplicate found
            if(rightMatchedIndex != -1){
                rightIndex = rightMatchedIndex;
                //increment lower bound of right array
                rightLow = rightMatchedIndex + 1;
            } else{
                break;
            }

        }

        return new Matches(matchedIndex, rightIndex - leftIndex + 1);

    }

    return new Matches(-1, 0);

}

private static int search(int[] values, int low, int high, int query) {

    while (low <= high) {
        int mid = (low + high) / 2;

        if (values[mid] == query) {
            return mid;
        } else if (query < values[mid]) {
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }

    return -1;

}

Answer 4

在更正导致无限循环的高变量重置错误后，我找到了解决方案。

public static Matches findMatches(int[] values, int query) {
    int firstMatchIndex = -1;
    int lastMatchIndex = -1;
    int numberOfMatches = 0;

    int low = 0;
    int mid = 0;
    int high = values.length - 1;

    while (low <= high){
        mid = (low + high)/2;

        if (values[mid] == query && firstMatchIndex == -1){

            firstMatchIndex = mid;
            numberOfMatches++;
            high = values.length - 1;
            low = mid;

        } else if (values[mid] == query && (lastMatchIndex == -1 || lastMatchIndex != -1)){

            lastMatchIndex = mid;
            numberOfMatches++;

            if (query < values[mid]){
                high = mid - 1;
            } else { 
                low = mid + 1;
            }

        } else if (query < values[mid]){
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }

    if (firstMatchIndex != -1) { // First match index is set
        return new Matches(firstMatchIndex, numberOfMatches);
    }
    else { // First match index is not set
        return new Matches(-1, 0); 
    }
}

Answer 5

除了先验排序以外，对数据一无所知是很困难的。 请参阅以下内容： 二进制搜索O（log n）算法在顺序列表中查找重复项？

这将在排序数组中找到k的重复项的第一个索引。 当然，这与首先知道重复项的值有关，但在知道重复项的值时非常有用。

    public static int searchFirstIndexOfK(int[] A, int k) {

     int left = 0, right = A.length - 1, result = -1;
     // [left : right] is the candidate set.
     while (left <= right) {
       int mid = left + ((right - left) >>> 1); // left + right >>> 1;
       if (A[mid] > k) {
         right = mid - 1;
       } else if (A[mid] == k) {
         result = mid;
         right = mid - 1; // Nothing to the right of mid can be
                                               // solution.
      } else { // A[mid] < k
      left = mid + 1;
      }
     }
     return result;
    }

这会发现log（n）时间上的重复，但是它很脆弱，因为必须对数据进行排序并将其增加1，且范围为1..n。

static int findeDupe(int[] array) {
int low = 0;
int high = array.length - 1;
while (low <= high) {
    int mid = (low + high) >>> 1;
    if (array[mid] == mid) {
    low = mid + 1;

    } else {
    high = mid - 1;

    }

}
System.out.println("returning" + high);
return high;

}