[leetcode] Search for a Range

最新推荐文章于 2018-09-05 14:58:19 发布

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本文介绍了一个算法问题：在已排序的整数数组中查找给定目标值的起始和结束位置，若未找到则返回[-1,-1]。该算法的时间复杂度为O(log n)，并提供了一个具体的C++实现案例。

From : https://leetcode.com/problems/search-for-a-range/

Given a sorted array of integers, 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].

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Array Binary Search

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(M) Search Insert Position

Solution :

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int start=0, end=nums.size()-1;
        vector<int> ans(2);
        ans[0]=ans[1]=-1;
        while(start <= end) {
            int mid = (start+end)>>1;
            if(nums[mid] == target) {
                for(int i=mid; i>=start; i--) {
                    if(nums[i] == target) {
                        ans[0] = i;
                    }
                }
                for(int i=mid; i<=end; i++) {
                    if(nums[i] == target) {
                        ans[1] = i;
                    }
                }
                break;
            }
            if(nums[mid] > target) {
                end = mid-1;
            } else {
                start = mid+1;
            }
        }
        return ans;
    }
};