LeetCode Search for a Range

题目：

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].

class Solution {public:vector<int> searchRange(int A[], int n, int target) {vector<int> ans(2, -1);ans[0] = findLeft(A, n, target);ans[1] = findRight(A, n, target);return ans;}private:int findLeft(int A[], int n, int target) {int left = -1;int low = 0, high = n - 1;while (low <= high) {int mid = (low + high) / 2;if (A[mid] < target) {low = mid+1;}else {if (A[mid] == target)left = mid;high = mid-1;}}return left;}int findRight(int A[], int n, int target) {int right = -1;int low = 0, high = n - 1;while (low <= high) {int mid = (low + high) / 2;if (A[mid] <= target) {if (A[mid] == target)right = mid;low = mid + 1;}elsehigh = mid-1;}return right;}};