LeetCode34 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].  （Medium）

分析：

标准的二分搜索题，找一个target的范围，也就是first position 元素等于target 和last position元素等于。

写两次二分搜索，注意中间start，end的变化情况的不同，一个为了保留住第一个满足条件的，一个为了保留住最后一个满足条件的。

代码：

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