我试图通过重置高变量来扩展函数以通过二进制搜索找到整数匹配的数量，但它会陷入循环中。我猜测一个解决方法是复制此函数以获取最后一个索引以确定匹配数，但我不认为这将是一个如此优雅的解决方案。
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);
}
}
答案 0 :(得分：3)
您的代码存在问题：
high = values[values.length - 1];
应该是
high = values.length - 1;
此外，您不需要像numberOfMatches和searchFirst这样的变量，我们可以有相当简单的解决方案。
现在遇到问题，我明白你想要什么我认为二进制搜索适合这种查询。
执行所需操作的最佳方法是一旦找到匹配项，您只需从该索引前进和后退，直到发生不匹配，这在计算firstMatchIndex和numberOfMatches时既优雅又高效。
所以你的功能应该是：
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);
}
答案 1 :(得分：0)
难道你不能只使用类似集合来查找重复项吗？
这样的事情：
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
答案 2 :(得分：0)
我将您的问题分成两部分 - 使用二分搜索来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;
}
答案 3 :(得分：0)
我通过重置导致无限循环的高变量来纠正错误后找到了解决方案。
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);
}
}
答案 4 :(得分：0)
除了先验分类之外，对数据没有任何了解是很困难的。 看到这个： Binary Search O(log n) algorithm to find duplicate in sequential list?
这将在排序数组中找到k的重复的第一个索引。 当然，这与首先知道重复的值有关，但在知道副本时非常有用。
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，但是很脆弱，因为必须对数据进行排序以及增加1并且在1..n范围内。
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;
}