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[英]How to find number of integers in a sorted array that are within a certain range in O(log(N)) time?
[英]Efficient algo to find number of integers in a sorted array that are within a certain range in O(log(N)) time?
我遇到了一個必須在O(logn)中完成的面試問題
給定排序的整數數組和數字,找到數組中數字的開始和結束索引。
Ex1: Array = {0,0,2,3,3,3,3,4,7,7,9} and Number = 3 --> Output = {3,6}
Ex2: Array = {0,0,2,3,3,3,3,4,7,7,9} and Number = 5 --> Output = {-1,-1}
我試圖找到一個有效的算法,但這么胖也沒有成功。
您可以使用二進制搜索的概念來查找起始和結束索引:
請注意,當我們到達大小為1的數組時,我們可能是輸入數字旁邊的一個單元格,因此我們檢查它是否等於輸入數字,如果不是,我們通過從我們找到的索引中添加/減少1來修復索引。
findStartIndex(int[] A, int num)
{
int start = 0; end = A.length-1;
while (end != start)
{
mid = (end - start)/2;
if (A[mid] >= num)
end = mid;
else
start = mid;
}
if(A[start] == num)
return start;
else
return start+1;
}
findEndIndex(int[] A, int num)
{
int start = 0; end = A.length-1;
while (end != start)
{
mid = (end - start)/2;
if (A[mid] > num)
end = mid;
else
start = mid;
}
if(A[start] == num)
return start;
else
return start-1;
}
整個過程:
int start = findStartIndex(A, num);
if (A[start]!=num)
{
print("-1,-1");
}
else
{
int end = findEndIndex(A, num);
print(start, end);
}
聽起來像二進制搜索 - 日志圖表iirc表示每個增量“減半”的效果,基本上是二進制搜索。
偽代碼:
Set number to search for
Get length of array, check if number is at the half point
if the half is > the #, check the half of the bottom half. is <, do the inverse
repeat
if the half point is the #, mark the first time this happens as a variable storing its index
then repeat binary searches above , and then binary searches below (separately), such that you check for how far to the left/right it can repeat.
note*: and you sort binary left/right instead of just incrementally, in case your code is tested in a dataset with like 1,000,000 3's in a row or something
這是否足夠清楚?
解決方案是在開始時同時二進制搜索數組(實際上不必是並發:P)。 關鍵是左右搜索略有不同。 對於右側,如果遇到欺騙,則必須向右搜索,對於左側,如果遇到欺騙,則向左搜索。 您要搜索的是邊界,所以在右側檢查。
yournum, not_yournum
這是邊界,在左側,您只需搜索相反方向的邊界。 最后返回邊界的索引。
雙二進制搜索。 您從較低的索引= 0開始,較高的索引=長度 - 1.然后您檢查點中途並相應地調整您的索引。
訣竅在於,一旦找到目標,樞軸就會分成兩個樞軸。
由於還沒有人發布工作代碼,我將發布一些(Java):
public class DuplicateNumberRangeFinder {
public static void main(String[] args) {
int[] nums = { 0, 0, 2, 3, 3, 3, 3, 4, 7, 7, 9 };
Range range = findDuplicateNumberRange(nums, 3);
System.out.println(range);
}
public static Range findDuplicateNumberRange(int[] nums, int toFind) {
Range notFound = new Range(-1, -1);
if (nums == null || nums.length == 0) {
return notFound;
}
int startIndex = notFound.startIndex;
int endIndex = notFound.endIndex;
int n = nums.length;
int low = 0;
int high = n - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == toFind && (mid == 0 || nums[mid - 1] < toFind)) {
startIndex = mid;
break;
} else if (nums[mid] < toFind) {
low = mid + 1;
} else if (nums[mid] >= toFind) {
high = mid - 1;
}
}
low = 0;
high = n - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == toFind && (mid == n - 1 || nums[mid + 1] > toFind)) {
endIndex = mid;
break;
} else if (nums[mid] <= toFind) {
low = mid + 1;
} else if (nums[mid] > toFind) {
high = mid - 1;
}
}
return new Range(startIndex, endIndex);
}
private static class Range {
int startIndex;
int endIndex;
public Range(int startIndex, int endIndex) {
this.startIndex = startIndex;
this.endIndex = endIndex;
}
public String toString() {
return "[" + this.startIndex + ", " + this.endIndex + "]";
}
}
}
這可能是我的錯誤,但是當我測試它時,Ron Teller的答案有一個無限循環。 這是Java中的一個工作示例,如果將searchRange
函數更改為不是靜態的,可以在此處進行測試。
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class RangeInArray {
// DO NOT MODIFY THE LIST
public static ArrayList<Integer> searchRange(final List<Integer> a, int b) {
ArrayList<Integer> range = new ArrayList<>();
int startIndex = findStartIndex(a, b);
if(a.get(startIndex) != b) {
range.add(-1);
range.add(-1);
return range;
}
range.add(startIndex);
range.add(findEndIndex(a, b));
return range;
}
public static int findStartIndex(List<Integer> a, int b) {
int midIndex = 0, lowerBound = 0, upperBound = a.size() - 1;
while(lowerBound < upperBound) {
midIndex = (upperBound + lowerBound) / 2;
if(b <= a.get(midIndex)) upperBound = midIndex - 1;
else lowerBound = midIndex + 1;
}
if(a.get(lowerBound) == b) return lowerBound;
return lowerBound + 1;
}
public static int findEndIndex(List<Integer> a, int b) {
int midIndex = 0, lowerBound = 0, upperBound = a.size() - 1;
while(lowerBound < upperBound) {
midIndex = (upperBound + lowerBound) / 2;
if(b < a.get(midIndex)) upperBound = midIndex - 1;
else lowerBound = midIndex + 1;
}
if(a.get(lowerBound) == b) return lowerBound;
return lowerBound - 1;
}
public static void main(String[] args) {
ArrayList<Integer> list = new ArrayList<>();
list.add(1);
list.add(1);
list.add(2);
list.add(2);
list.add(2);
list.add(2);
list.add(2);
list.add(2);
list.add(3);
list.add(4);
list.add(4);
list.add(4);
list.add(4);
list.add(5);
list.add(5);
list.add(5);
System.out.println("Calling search range");
for(int n : searchRange(list, 2)) {
System.out.print(n + " ");
}
}
}
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