[LeetCode]Search for a Range

题目描述:(链接)

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].

解题思路:

二分查找

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result;
        int len = nums.size();
        int first = 0;
        int last = len - 1;
        while (first <= last) {
            int mid = first + (last - first) / 2;
            if (nums[mid] == target) {
                int l1 = mid;
                int l2 = mid;
                while (--l1 >= 0 && nums[l1] == target);
                while (++l2 < len && nums[l2] == target);
                result.push_back(++l1);
                result.push_back(--l2);
                break;
            } else if (target < nums[mid]) {
                last = mid - 1;
            } else {
                first = mid + 1;
            }
        }
        
        if (result.size() == 0) {
            result.push_back(-1);
            result.push_back(-1);
        }
        
        return result;
    }
};