1. Knowledge points of this question
Binary search
2. Title description
Count the number of times a number appears in the ascending array.
输入：
[1,2,3,3,3,3,4,5],3
返回：
4
3. Problem solving ideas
Use dichotomy to find the first subscript where the repeated element appears, then use dichotomy to find the last subscript where the repeated element appears, and subtract to get the number of occurrences.
4. Code
public class Solution {
public int GetNumberOfK(int[] array, int k) {
int firstIndex = GetFirstIndex(array, k);
int lastIndex = GetLastIndex(array, k);
if (firstIndex == -1 || lastIndex == -1) {
return 0;
}
return lastIndex - firstIndex + 1;
}
public int GetFirstIndex(int[] array, int k) {
int high = array.length - 1;
int low = 0;
while (low <= high) {
int mid = (low + high) / 2;
if (k > array[mid]) {
// 如果 k > 中间值，则在右半数组查找
low = mid + 1;
} else if (k < array[mid]) {
// 如果 k < 中间值，则在左半数组查找
high = mid - 1;
} else {
// 如果 k = 中间值，则判断 mid-1 是否等于 k
mid = mid - 1;
if (mid < low || array[mid] != k) {
// 不等于 k，则找到第一个下标
return mid + 1;
} else {
// 等于 k,继续二分查找
high = mid;
}
}
}
return -1;
}
public int GetLastIndex(int[] array, int k) {
int high = array.length - 1;
int low = 0;
while (low <= high) {
int mid = (low + high) / 2;
if (k > array[mid]) {
low = mid + 1;
} else if (k < array[mid]) {
high = mid - 1;
} else {
mid = mid + 1;
if (mid > high || array[mid] != k) {
return mid - 1;
} else {
low = mid;
}
}
}
return -1;
}
}