[LeetCode] 34. Search for a Range 搜索一个范围(Find First and Last Position of Element in Sorted Array)

原题目：Search for a Range， 现在题目改为： 34. 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:

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]

给一个有序整数数组中，寻找相同目标值的起始和结束位置，限定了时间复杂度为O(logn)。

解法：二分法，典型的二分查找法的时间复杂度，先对原数组使用二分查找法，找出其中一个目标值的位置，然后向两边搜索找出起始和结束的位置。

Java:

public class Solution {
	public int[] searchRange(int[] A, int target) {
		int start = Solution.firstGreaterEqual(A, target);
		if (start == A.length || A[start] != target) {
			return new int[]{-1, -1};
		}
		return new int[]{start, Solution.firstGreaterEqual(A, target + 1) - 1};
	}

	//find the first number that is greater than or equal to target.
	//could return A.length if target is greater than A[A.length-1].
	//actually this is the same as lower_bound in C++ STL.
	private static int firstGreaterEqual(int[] A, int target) {
		int low = 0, high = A.length;
		while (low < high) {
			int mid = low + ((high - low) >> 1);
			//low <= mid < high
			if (A[mid] < target) {
				low = mid + 1;
			} else {
				//should not be mid-1 when A[mid]==target.
				//could be mid even if A[mid]>target because mid<high.
				high = mid;
			}
		}
		return low;
	}
}　　

Python:

class Solution(object):
    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        # Find the first idx where nums[idx] >= target
        left = self.binarySearch(lambda x, y: x >= y, nums, target)
        if left >= len(nums) or nums[left] != target:
            return [-1, -1]
        # Find the first idx where nums[idx] > target
        right = self.binarySearch(lambda x, y: x > y, nums, target)
        return [left, right - 1]

    def binarySearch(self, compare, nums, target):
        left, right = 0, len(nums)
        while left < right:
            mid = left + (right - left) / 2
            if compare(nums[mid], target):
                right = mid
            else:
                left = mid + 1
        return left

    def binarySearch2(self, compare, nums, target):
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = left + (right - left) / 2
            if compare(nums[mid], target):
                right = mid - 1
            else:
                left = mid + 1
        return left

    def binarySearch3(self, compare, nums, target):
        left, right = -1, len(nums)
        while left + 1 < right:
            mid = left + (right - left) / 2
            if compare(nums[mid], target):
                right = mid
            else:
                left = mid
        return left if left != -1 and compare(nums[left], target) else right

C++:

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        const auto start = lower_bound(nums.cbegin(), nums.cend(), target);
        const auto end = upper_bound(nums.cbegin(), nums.cend(), target);
        if (start != nums.cend() && *start == target) {
            return {start - nums.cbegin(), end - nums.cbegin() - 1};
        }
        return {-1, -1};
    }
};

class Solution2 {
public:
    vector<int> searchRange(vector<int> &nums, int target) {
        const int begin = lower_bound(nums, target);
        const int end = upper_bound(nums, target);

        if (begin < nums.size() && nums[begin] == target) {
            return {begin, end - 1};
        }

        return {-1, -1};
    }

private:
    int lower_bound(vector<int> &nums, int target) {
        int left = 0;
        int right = nums.size();
        // Find min left s.t. A[left] >= target.
        while (left < right) {
            const auto mid = left + (right - left) / 2;
            if (nums[mid] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    int upper_bound(vector<int> &nums, int target) {
        int left = 0;
        int right = nums.size();
        // Find min left s.t. A[left] > target.
        while (left < right) {
            const auto mid = left + (right - left) / 2;
            if (nums[mid] > target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

原文地址：https://www.cnblogs.com/lightwindy/p/9667030.html