Leetcode: 34. Search for a Range

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].

Example

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

思路

• 二分查找
• 两次二分，分别查找左边起点和右边终点
• 对应于代码中flag = 1,找左边；flag = 0, 找右边

代码

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
		vector<int> res;
		int len = nums.size();
		
		res.push_back(binarySearch(nums, len, target, 1));
		res.push_back(binarySearch(nums, len, target, 0));

		return res;
	}

	int binarySearch(vector<int>& nums, int len, int target, int flag){
		int low = 0, high = len - 1;
		int mid = 0;
		while (low <= high){
			mid = low + (high - low) / 2;

			if (nums[mid] == target){
				if (flag){
					if (mid == low || mid - 1 >= low && nums[mid - 1] < nums[mid])
						return mid;
					else high = mid - 1;
				}
				else{
					if (mid == high || mid + 1 <= high && nums[mid + 1] > nums[mid])
						return mid;
					else low = mid + 1;
				}
			}
			else if (nums[mid] > target)
				high = mid - 1;
			else low = mid + 1;
		}

		return -1;
	}
};