import java.util.Arrays;
/**
* Source : https://oj.leetcode.com/problems/search-for-a-range/
*
* Created by lverpeng on 2017/7/14.
*
* Given a sorted array of integers, find the starting and ending position of a given target value.
*
* Your algorithm's runtime complexity must be in the order of O(log n).
*
* If the target is not found in the array, return [-1, -1].
*
* For example,
* Given [5, 7, 7, 8, 8, 10] and target value 8,
* return [3, 4].
*
*/
public class SearchRange {
/**
* 查找target在有有序数组中的起始位置
*
* 先找左边界,普通二分查找是和target比较,如果相同就返回,这里小于等于num[mid],如果是等于num[mid]也是收缩右边,最后得到的就是左边界
* 右边界同上
*
* @param num
* @param target
* @return
*/
public int[] search (int[] num, int target) {
int left = 0;
int right = num.length - 1;
int mid = 0;
while (left <= right) {
mid = (left + right) / 2;
if (num[mid] >= target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
int targetLeft = left;
left = 0;
right = num.length - 1;
while (left <= right) {
mid = (left + right) / 2;
if (num[mid] <= target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
int targetRight = right;
if (target != num[targetLeft] || target != num[targetRight]) {
targetLeft = targetRight = -1;
}
int[] result = new int[2];
result[0] = targetLeft;
result[1] = targetRight;
return result;
}
public static void main(String[] args) {
SearchRange searchRange = new SearchRange();
int[] arr = new int[]{5, 7, 7, 8, 8, 10};
System.out.println(Arrays.toString(searchRange.search(arr, 8)));
}
}