LeetCode Search for a Range

最新推荐文章于 2018-09-05 14:58:19 发布

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本文介绍了一种在有序整数数组中查找指定目标值起始和结束位置的算法。该算法需满足O(log n)的时间复杂度要求。如果未找到目标值，则返回[-1,-1]。举例说明了当输入为[5,7,7,8,8,10]且目标值为8时，算法返回[3,4]。

题目：

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].

class Solution {
public:
	vector<int> searchRange(int A[], int n, int target) {
		vector<int> ans(2, -1);
		ans[0] = findLeft(A, n, target);
		ans[1] = findRight(A, n, target);
		return ans;
	}
private:
	int findLeft(int A[], int n, int target) {
		int left = -1;
		int low = 0, high = n - 1;
		while (low <= high) {
			int mid = (low + high) / 2;
			if (A[mid] < target) {
				low = mid+1;
			}
			else {
				if (A[mid] == target)
					left = mid;
				high = mid-1;	
			}
		}
		return left;
	}
	int findRight(int A[], int n, int target) {
		int right = -1;
		int low = 0, high = n - 1;
		while (low <= high) {
			int mid = (low + high) / 2;
			if (A[mid] <= target) {
				if (A[mid] == target)
					right = mid;
				low = mid + 1;

			}
			else
				high = mid-1;
		}
		return right;
	}
};