leetcode 034 Search for a Range

最新推荐文章于 2018-06-21 15:37:28 发布

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本文介绍了一个算法，该算法能在已排序的整数数组中找到指定目标值的起始和结束位置，采用二分查找实现O(log n)的时间复杂度。

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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#include <iostream>
#include <cstdio>
#include <string>
using namespace std;

class Solution {
public:
	vector<int> searchRange(vector<int>& nums, int target) {
		int l = 0, r = nums.size(), pos=-1, left=-1, right=-1;
		vector<int> vec;

		while(l <= r) {
			int mid = (l+r)/2;

			if(nums[mid] == target) {
				pos = mid;
				break;
			}
			else if(nums[mid] > target) r = mid-1;

			else l = mid+1;
		}

		if(l > r) {

			vec.push_back(left);
			vec.push_back(right);

			return vec;
		}

		l = 0;
		r = pos-1;

		while(l <= r && nums[r] >= target) {
			int mid = (l+r)/2;
			if(nums[mid] < target) l = mid+1;
			else if(nums[mid] == target) r = mid-1; 
		}

		left = r+1, right=-1;

		l = pos+1;
		r = nums.size()-1;

		while(l <= r && nums[l] <= target) {
			int mid = (l+r)/2;
			if(nums[mid] == target) l=mid+1;
			else if(nums[mid] > target) r=mid-1;
		}

		right = l-1;

		vec.push_back(left);
		vec.push_back(right);

		return vec;
	}
}