[LeetCode] 34. Search for a Range

題目鏈接: 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].

解題思路

題意爲給定已排序數組，以及一個目標值，找出目標值在數組中的下標範圍。因爲是已排序數組，所以馬上想到的就是用二分查找，通過兩次二分查找確定範圍，一次用來找目標值在數組中第一次出現的下標，一次用來找目標值在數組中最後一次出現的下標。其中，前者用的是找第一個大於或等於目標值的二分查找算法，後者用的時找最後一個小於或等於目標值的二分查找算法，若第一次二分查找找到的值不是目標值，就說明數組中沒有該值，直接返回 [-1, -1]，不用進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;
    }
};