leetcode 34. Search for a Range

题目描述

找到一个升序序列中某个数在哪个范围内，没有的话返回【-1，-1】
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].

代码实现

使用二分法找到目标数字的位置A，然后往A的前面和后面走找到边界。

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int nums_len = nums.size();        int lo = 0, hi = nums_len - 1;        vector<int> result;        int mid = 0;        if(nums_len < 1) {            result.push_back(-1);            result.push_back(-1);                        return result;        }            while(lo <= hi) {            mid = (lo + hi) >> 1;            if(nums[mid] > target)                hi = mid - 1;            else if(nums[mid] < target)                lo = mid + 1;            else                break;        }        int tmp = mid;        int left = -1, right = -1;        while(mid >= 0 && nums[mid] == target) {            left = mid;            mid--;        }        while(tmp < nums_len && nums[tmp] == target) {            right = tmp;            tmp++;        }        result.push_back(left);        result.push_back(right);                    return result;    }};