題目

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].

解題之法

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

分析

這道題讓我們在一個有序整數數組中尋找相同目標值的起始和結束位置，而且限定了時間復雜度為O(logn)，這是典型的二分查找法的時間復雜度，所以這道題我們也需要用此方法。
首先對原數組使用二分查找法，找出其中一個目標值的位置，然后向兩邊搜索找出起始和結束的位置