leetcode解法: Search for a Range (34)

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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].

解法1：Binary Search, O(logN)

看到在sorted array里找元素的问题，首先想到可以用binary search。 这里用两次binarysearch， 一次找到range的起始点，一次找到最后一个点。
C++

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

Java

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