LeetCode Search for a Range

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].

题意：求已排序数组出现某个数的范围。

思路：两次二分，找出左右边界

class Solution {public:vector<int> searchRange(int A[], int n, int target) {int left = 0, right = n;while (left < right) {int mid = left + right >> 1;if (A[mid] >= target) right = mid;else left = mid + 1;}int ansLeft = left;left = 0, right = n;while (left < right) {int mid = left + right >> 1;if (A[mid] > target)right = mid;else left = mid + 1;}if (A[ansLeft] != target)return vector<int>{-1, -1};else return vector<int>{ansLeft, left-1};}};