Leetcode 34. Search for a Range

题目链接

问题描述

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:

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

Example 2:

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

解题思路

使用两次二分查找，分别找出上界和下界，如果nums[start]==target，则下界索引就是start，否则为-1，上界求解方法类似。

代码如下

class Solution {
public:
	vector<int> searchRange(vector<int>& nums, int target) {
		vector<int> ret;
		if (nums.size() == 0 || nums[0] > target || nums[nums.size() - 1] < target) {
			ret.push_back(-1);
			ret.push_back(-1);
			return ret;
		}
		int leng = nums.size();
		int start = 0, end = leng - 1;
		int mid;

		//寻找左边边界
		while (start < end) {
			mid = (end - start) / 2 + start;
			if (nums[mid] < target) start = mid + 1;
			else end = mid;
		}
		if (nums[start] == target) ret.push_back(start);
		else ret.push_back(-1);

		//寻找右边边界
		start = 0, end = leng - 1;
		while (start < end) {
			mid = (end - start + 1) / 2 + start;
			//此处若跟寻找左边边界的代码一样,则当nums[mid]<=target时，若start+1==end就会陷入死循环
			if (nums[mid] > target) end = mid - 1;
			else start = mid;
		}
		if (nums[start] == target) ret.push_back(start);
		else ret.push_back(-1);
		return ret;
	}
};