leetcode 34. Search for a Range

二分查找算法在找到目标值后的边界处理
本文详细介绍了如何使用二分查找算法在已排序的整数数组中找到给定目标值的起始和结束位置。通过示例说明了在找到目标值后,如何进一步扩展查找范围以确定其边界。

1.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].

2.思路 

二分查找,然后向两边延伸。。。

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


### LeetCode Problem 34: Find First and Last Position of Element in Sorted Array The task involves finding the starting and ending position of a given target value within an array of integers. If the target is not found in the array, [-1, -1] should be returned. For instance, consider an input where `nums` = [5,7,7,8,8,10], and `target` = 8. The expected output would be [3, 4]. Another example could involve `nums` = [5,7,7,8,8,10], but this time with `target` = 6, leading to an output of [-1, -1]. To solve this problem efficiently: A binary search approach can achieve logarithmic complexity by narrowing down potential positions for both the first and last occurrence of the target element[^1]: ```python def searchRange(nums, target): def findLeftIndex(nums, target): left, right = 0, len(nums) - 1 while left <= right: mid = (left + right) // 2 if nums[mid] < target: left = mid + 1 else: right = mid - 1 return left def findRightIndex(nums, target): left, right = 0, len(nums) - 1 while left <= right: mid = (left + right) // 2 if nums[mid] <= target: left = mid + 1 else: right = mid - 1 return right left_index = findLeftIndex(nums, target) right_index = findRightIndex(nums, target) # Check if the target exists in the list. if left_index <= right_index < len(nums) and nums[left_index] == target: return [left_index, right_index] return [-1, -1] ``` This code snippet defines two helper functions that perform modified versions of binary searches—one looking for the start index (`findLeftIndex`) and another for the end index (`findRightIndex`). After determining these indices, it checks whether they are valid before returning them as part of the result.
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