如何使用二進制搜索在排序數組中查找重復項？

Question

我試圖通過重置高變量來擴展通過二進制搜索找到整數匹配數的函數，但是它陷入了循環。 我猜一種解決方法是復制此函數以獲得最后一個索引來確定匹配數，但是我認為這不是一個很好的解決方案。

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); 
    }
}

對於這樣的事情：

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

您的代碼有問題：

high = values[values.length - 1];

應該

high = values.length - 1;

另外，您不需要numberOfMatches和searchFirst之類的變量，我們可以有一個非常簡單的解決方案。

現在來解決這個問題，我了解您想要什么，我認為二進制搜索適合這種查詢。

達到要求的最佳方法是， 一旦找到匹配項，您就從該索引向前或向后前進，直到發生不匹配為止，這在計算firstMatchIndex和numberOfMatches方面既優雅又有效。

因此，您的功能應為：

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來查找重復項嗎？

像這樣：

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;
    }

}

輸出：

DUPLICATES:
    two
    six

Answer 3

我將您的問題分為兩部分-使用二進制搜索查找數字並計算匹配數。 第一部分由搜索功能解決，而第二部分由findMatches函數解決：

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

在更正導致無限循環的高變量重置錯誤后，我找到了解決方案。

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

除了先驗排序以外，對數據一無所知是很困難的。 請參閱以下內容： 二進制搜索O（log n）算法在順序列表中查找重復項？

這將在排序數組中找到k的重復項的第一個索引。 當然，這與首先知道重復項的值有關，但在知道重復項的值時非常有用。

    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;
    }

這會發現log（n）時間上的重復，但是它很脆弱，因為必須對數據進行排序並將其增加1，且范圍為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;

}