java查找排序数组中重复值的计数

Question

我只是想知道这段代码是否可以进一步改进。这是一个简单的类，它接受排序数组和值作为参数，并尝试查找该值在数组中出现的次数

据我所知，代码的复杂度是O（logn+k）[假设k是n的子集]。这个代码可以进一步改进吗

public class SearchSortedArray {
// Find number count in an array. Total Complexity = O(log n + k) [assuming k is the subset of n] 
public static int findNumberCountInAnArray(List<Integer> array, Integer findValue) {
    //Search Index
    Integer searchIndex = findIndex(array, findValue, 0);
    //Find Count
    return countArray(array, searchIndex);
}

// Search Index. Complexity = O(log n)
private static Integer findIndex(List<Integer> array, Integer findValue, int offset) {
    if(array.size() == 0) {
        return null;
    }
    if(findValue < array.get(array.size()/2)) {
        return findIndex(array.subList(0, array.size()/2), findValue, 0);
    } else if(findValue > array.get(array.size()/2)) {
        offset = offset + array.size()/2;
        return findIndex(array.subList(array.size()/2, array.size()), findValue,  offset);
    }
    return array.size()/2+offset;           
}

// Find count. Complexity = O(k) [assuming k is the subset of n] 
private static int countArray(List<Integer> array, Integer searchIndex) {
    if(null == searchIndex) {
        return 0;
    }
    Integer searchValue = array.get(searchIndex);
    Integer searchIndexStore = searchIndex;
    int count = 0;
    while(searchIndex < array.size() && searchValue == array.get(searchIndex)) {
        count++;
        searchIndex++;
    }
    searchIndex = searchIndexStore;
    while(searchIndex > 0 && searchValue == array.get(searchIndex-1)) {
        count++;
        searchIndex--;
    }
    return count;
}
}

这是主课

// Test main class. 
public class TestMain {
public static void main(String[] args) {
    Integer[] sample1 = {1, 1, 1, 2, 3, 4, 4, 4, 4, 5, 5, 5}; 
    ArrayList<Integer> arraySample1 = new ArrayList<Integer>(Arrays.asList(sample1));
    // Find the number of times 5 is repeated in the array?
    System.out.println(SearchSortedArray.findNumberCountInAnArray(arraySample1, 5));

    Integer[] sample2 = {1, 1, 2, 3, 3, 4, 5, 8, 8, 9, 9, 10, 10, 14, 18};
    ArrayList<Integer> arraySample2 = new ArrayList<Integer>(Arrays.asList(sample2));   
    // Find the number of times 10 is repeated in the array?
    System.out.println(SearchSortedArray.findNumberCountInAnArray(arraySample2, 10));
  }
}

谢谢

Answer 1

使用两种方法代替一种二进制搜索：

1. 首先查找值findValue的第一次出现。让我们称之为索引lo。我相信这就是你的findIndex函数所做的
2. 然后查找第一次出现的大于findValue的值，或者n如果不存在这样的值。让我们称之为索引hi

现在结果是hi - lo

运行时：O（日志n）。请注意，如果k的预期值很小，即在log（n）的小常数因子内，那么您的代码实际上可能会更好，因为线性搜索比二进制搜索更高效