Search for a Range

描述​

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].

重新实现 lower_bound 和 upper_bound​

// Search for a Range// 重新实现 lower_bound 和 upper_bound// 时间复杂度O(logn)，空间复杂度O(1)public class Solution {    public int[] searchRange(int[] nums, int target) {        int lower = lower_bound(nums, 0, nums.length, target);        int upper = upper_bound(nums, 0, nums.length, target);        if (lower == nums.length || nums[lower] != target)            return new int[]{-1, -1};        else            return new int[]{lower, upper-1};    }    int lower_bound (int[] A, int first, int last, int target) {        while (first != last) {            int mid = first + (last - first) / 2;            if (target > A[mid]) first = ++mid;            else                 last = mid;        }        return first;    }    int upper_bound (int[] A, int first, int last, int target) {        while (first != last) {            int mid = first + (last - first) / 2;            if (target >= A[mid]) first = ++mid;  // 与 lower_bound 仅此不同            else                  last = mid;        }        return first;    }}