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NumOfK.java
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package Chap6;
/**
* 统计一个数字在排序数组中出现的次数。
*/
public class NumOfK {
/**
* 方法一：遍历，O(n)复杂度，不推荐
*/
public int getNumberOfK(int[] array, int k) {
if (array == null) return 0;
int count = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == k) {
count++;
}
}
return count;
}
/**
* 方法二：二分法找到第一个k和最后一个k，时间复杂度O(nlgn)
*/
public int numberOfK(int[] array, int k) {
if (array == null) return 0;
int from = getFirstOfK(array, k, 0, array.length - 1);
int to = getLastOfK(array, k, 0, array.length - 1);
if (from == -1 && to == -1) return 0;
else return to - from + 1;
}
private int getFirstOfK(int[] array, int k, int low, int high) {
while (low <= high) {
int mid = low + (high - low) / 2;
if (k < array[mid]) high = mid - 1;
else if (k > array[mid]) low = mid + 1;
else {
if (mid > 0 && array[mid - 1] == k) high = mid - 1;
else return mid;
}
}
return -1;
}
private int getLastOfK(int[] array, int k, int low, int high) {
while (low <= high) {
int mid = low + (high - low) / 2;
if (k < array[mid]) high = mid - 1;
else if (k > array[mid]) low = mid + 1;
else {
if (mid < array.length - 1 && array[mid + 1] == k) low = mid + 1;
else return mid;
}
}
return -1;
}
/**
* 巧妙的方法3：得到与k相邻的两个浮点数排名
* 因为数组中元素时int型的,改为查找浮点型的值，比如找3的次数，就找2.5和3.5之间的元素个数，
* 稍微改变二分查找的返回值就能得到数组中排名为k的方法
*/
public int numOfK(int[] array, int k) {
if (array == null) return 0;
return rank(array, k + 0.5) - rank(array, k - 0.5);
}
// 因为数组中元素都是int型的,改为查找浮点型的值，比如要获得3的出现次数，就找2.5和3.5之间的元素个数，
// 稍微改变二分查找的返回值就能得到数组中排名为k的方法
private int rank(int[] array, double k) {
int low = 0;
int high = array.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (k < array[mid]) high = mid - 1;
else if (k > array[mid]) low = mid + 1;
}
return low;
}
public static void main(String[] args) {
int[] a = {1, 2, 3, 3, 3, 3, 4, 5};
NumOfK k = new NumOfK();
System.out.println(k.numOfK(a, 3));
}
}