【C++】【LeetCode】34. Search for a Range

题目

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].

思路

先用二分法找到某个target所在下标，然后往前后遍历，直到遍历到不同元素。

代码

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