LeetCode 34. Search for a Range

题目描述:

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].

先使用二分搜索查找到目标, 再向前后搜索到数值的起始位置.

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