[array] leetcode - 34. Search for a Range - Medium

leetcode - 34. Search for a Range - Medium

descrition

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].

解析

对于有序数组的查找问题，基本上都可以使用折半查找的思路。再者题目的要求复杂度是 O(log n)，因此我们需要两次折半查找。如果我们只用一次折半查找，找到数组 target 出现的任意一个位置，然后线性遍历找到最左边和最右，需要的复杂度是 O(n)。

具体实现代码如下。

code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution{
public:
	vector<int> searchRange(vector<int>& nums, int target) {
		vector<int> ans(2, -1);
		int ileft = binarySearchLeftMost(nums, target);
		if(ileft < 0)
			return ans;
		
		int iright = binarySearchRightMost(nums, target);

		ans[0] = ileft;
		ans[1] = iright;

		return ans;
	}

	int binarySearchLeftMost(vector<int>& nums, int target){
		int ans = -1; // save the left most index in nums which equal to target
		int ileft = 0, iright = nums.size() - 1;

		while(ileft <= iright){
			int imid = ileft + (iright - ileft) / 2;
			if(nums[imid] == target){
				ans = imid;
				iright = imid - 1;
			}else if (nums[imid] < target){
				ileft = imid + 1;
			}else{
				// nums[imid] > target
				iright = imid - 1;
			}
		}

		return ans;
	}

	int binarySearchRightMost(vector<int>& nums, int target){
		int ans = -1;
		int ileft = 0, iright = nums.size() - 1;

		while(ileft <= iright){
			int imid = ileft + (iright - ileft) / 2;
			if(nums[imid] == target){
				ans = imid;
				ileft = imid + 1;
			}else if (nums[imid] < target){
				ileft = imid + 1;
			}else {
				// nums[imid] > target
				iright = imid - 1;
			}
		}

		return ans;
	}


};

int main()
{
	return 0;
}