LeetCode 34. Search for a Range

最新推荐文章于 2024-03-22 00:51:13 发布

LchinaM

最新推荐文章于 2024-03-22 00:51:13 发布

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本文介绍了一种算法，可在已排序的整数数组中找到特定目标值的起始和结束位置，若未找到则返回[-1,-1]。该算法遵循O(log n)的时间复杂度要求。

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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].

answer:

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int head = -1,tail = -1,start = 0,end = nums.size() - 1;
        mySearch(nums,start,end,target,head,tail);
        vector<int> result;
        result.push_back(head);
        result.push_back(tail);
        return result;
    }
    
    void mySearch(vector<int>& nums, int start, int end, int target, int & head,int & tail){
        if(start > end) return;
        int mid = (start + end) / 2;
        if(nums[mid] < target){
            mySearch(nums,mid + 1,end,target,head,tail);
        }
        else if(nums[mid] > target){
            mySearch(nums,start,mid - 1,target,head,tail);
        }
        else{
            int i = mid;
            while(i < nums.size() && nums[i] == target) i++;
            if( i > mid) tail = i - 1;
            else tail = i;
            i = mid;
            while(i >= 0 && nums[i] == target) i--;
            if(i < mid) head = i + 1;
            else head = i;
            return;
        }
    }
};