34. Search for a Range

Given an array of integers sorted in ascending order, 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].

这一题的意思是给定一个以升序排好序的序列和一个目标值，然后找出该目标值的第一次出现的位置和最后一次出现的位置。题目比较简单，因为已经排好序了，所以一个O(logn)的做法就是用二分查找。所以二分查找也是比较基础的一个算法，不详细讲，代码如下：

Code(LeetCode运行9ms)

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> result(2);        result[0] = result[1] = -1;        if (nums.size() == 0) {            return result;        }        int pos = findTarget(nums, 0, nums.size() - 1, target);        result[0] = result[1] = pos;        for (int i = pos + 1; i < nums.size(); i++) {            if (nums[i] != target) {                result[1] = i - 1;                break;            } else {                result[1]++;            }        }        for (int i = pos - 1; i >= 0; i--) {            if (nums[i] != target) {                result[0] = i + 1;                break;            } else {                result[0]--;            }        }        return result;    }        int findTarget(vector<int> nums, int start, int end, int target) {        if (start > end) {            return -1;        }        int mid = start + (end - start) / 2;        if (nums[mid] == target) {            return mid;        } else if (nums[mid] < target) {            return findTarget(nums, mid + 1, end, target);        } else {            return findTarget(nums, start, mid - 1, target);        }    }};