34. Search for a Range

Given an array of integers nums 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].

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

找目标值涵盖的区间，要求时间复杂度O(logn)。

看到O(logn)，直接想到的就是二分查找。然后想到用两次二分法，分别求第一个下界和上界。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if (nums.empty()) 
            return {-1, -1};
        int l = -1;
        int r = nums.size();
        int mid, low, high;
        while (l < r - 1) {
            mid = l + (r - l) / 2;
            if (nums[mid] >= target)
                r = mid;
            else
                l = mid;
        }
        low = (r == nums.size() || nums[r] < target)? r: r-1;
        l = -1;
        r = nums.size();
        while (l < r - 1) {
            mid = l + (r - l) / 2;
            if (nums[mid] <= target)
                l = mid;
            else
                r = mid;
        }
        high = (l == -1 || nums[l] > target)? l: l+1;
        if (high - low > 1) {
            return {low+1, high-1};
        }
        return {-1, -1};
    }
};