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 high 变量来扩展一个函数以通过二进制搜索查找整数匹配的数量，但它陷入了循环。我猜一个解决方法是复制此函数以获得最后一个索引以确定匹配数，但我认为这不是一个优雅的解决方案。

来自这里：

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); 
    }
}

像这样：

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

你不能只使用像集合这样的东西来查找重复项吗？

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 2

您的代码有问题：

high = values[values.length - 1];

应该是

high = values.length - 1;

你也不需要像 numberOfMatches 和 searchFirst 这样的变量，我们可以有相当简单的解决方案。

现在进入问题，我明白你想要什么我认为二进制搜索适合这样的查询。

完成所需的最佳方法是一旦找到匹配项，您只需从该索引向前和向后移动直到出现不匹配，这在计算 firstMatchIndex 和 numberOfMatches 时既优雅又高效.

所以你的函数应该是：

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 3

我已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项数。第一部分由搜索函数解析，第二部分由 findMatches 函数解析：

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

在更正了导致死循环的重置高变量的错误后，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

除了先验排序之外，对数据一无所知是很困难的。看到这个： Binary Search O(log n) algorithm to find duplicate in sequential list?

这将找到排序数组中 k 重复项的第一个索引。当然，这与首先知道 duplicate 的值有关，但是当知道它时非常有用。

    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;
    }

这将在 log(n) 时间内找到一个欺骗，但它很脆弱，因为数据必须排序以及在 1..n. 范围内增加 1。

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;

}

如何使用二进制搜索在排序数组中查找重复项？

How to use Binary Search to find duplicates in sorted array?

java

sorted

binary-search