[LeetCode] 34. Search for a Range

题目链接: https://leetcode.com/problems/search-for-a-range/description/

Description

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].

解题思路

题意为给定已排序数组，以及一个目标值，找出目标值在数组中的下标范围。因为是已排序数组，所以马上想到的就是用二分查找，通过两次二分查找确定范围，一次用来找目标值在数组中第一次出现的下标，一次用来找目标值在数组中最后一次出现的下标。其中，前者用的是找第一个大于或等于目标值的二分查找算法，后者用的时找最后一个小于或等于目标值的二分查找算法，若第一次二分查找找到的值不是目标值，就说明数组中没有该值，直接返回 [-1, -1]，不用进行第二次二分查找了。

Code

class Solution {public:    vector<int> searchRange(vector<int> &nums, int target) {        vector<int> res(2, -1);        if (nums.empty()) return res;        int fge = first_ge(nums, target);        if (fge == nums.size() || nums[fge] != target) return res;        int lle = last_le(nums, target);        res[0] = fge;        res[1] = lle;        return res;    }    int first_ge(vector<int> &input, int target) {        int left = 0, right = input.size() - 1;        while (left <= right) {            int middle = (left + right) >> 1;            if (target <= input[middle])                right = middle - 1;            else                left = middle + 1;        }        return left;    }    int last_le(vector<int> &input, int target) {        int left = 0, right = input.size() - 1;        while (left <= right) {            int middle = (left + right) >> 1;            if (target >= input[middle])                left = middle + 1;            else                right = middle - 1;        }        return right;    }};