Leetcode-34 Find First and Last Position of Element in Sorted Array

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Find First and Last Position of Element in Sorted Array

Given an array of integers nums 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 1

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Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2

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Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

解题思路

分别用二分法找出目标的左边界和右边界。找左边界时，若 nums[mid]>=target，二分搜索的上界 r=mid-1，否则二分搜索的下界 l=mid+1，最后返回 r+1。找右边界时，若 nums[mid]<=target，二分搜索的下界 l=mid+1，否则二分搜搜的上界 r=mid-1，最后返回 l-1。若左边界大于右边界，说明 nums 中不存在target，返回 [-1,-1]；否则，返回左右边界。

复杂度分析

  • 时间复杂度：$O(logn)$。
  • 空间复杂度：$O(1)$。

代码

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class Solution {
public:
	vector<int> searchRange(vector<int>& nums, int target) {
		int l = left(nums, target);
		int r = right(nums, target);
		if (l > r) return { -1,-1 };
		return { l,r };
	}
	int left(vector<int>& nums, int target) {
		int l = 0, r = nums.size() - 1;
		while (l <= r) {
			int mid = (l + r) / 2;
			if (nums[mid] < target)
				l = mid + 1;
			else r = mid - 1;
		}
		return r + 1;
	}
	int right(vector<int>& nums, int target) {
		int l = 0, r = nums.size() - 1;
		while (l <= r) {
			int mid = (l + r) / 2;
			if (nums[mid] > target)
				r = mid - 1;
			else l = mid + 1;
		}
		return l - 1;
	}
};