LeetCode Search for a Range-优快云博客

本文详细介绍了如何利用二分搜索算法，在已排序的整数数组中找到指定目标值的起始和结束索引。包括算法的复杂度分析和代码实现。

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题目要求：

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].

思路：可以用二分搜索分别找到最左端的位置和最右端的位置

代码：

class Solution {
public:
  vector<int> searchRange(int A[], int n, int target) 
  {
    int left = -1;
    int right = -1;
    vector<int> res;
    if(A == NULL || n <= 0)
    {
      res.push_back(left);
      res.push_back(right);
      return res;
    }
    left = FindLeftIndex(A, n, target);
    right = FindRightIndex(A, n, target);
    res.push_back(left);
    res.push_back(right);
    return res;
  }
  
  int FindLeftIndex(int A[], int n, int target)
  {
    int mid;
    int low = 0, high = n - 1;
    while(low <= high)
    {
      mid = low + ((high - low) >> 1);//防止溢出
      if(A[mid] == target && (mid == 0 || A[mid - 1] != target))
        return mid;
      if(A[mid] >= target)
        high = mid - 1;
      else 
        low = mid + 1;
    }
    return -1;
  }
  
  int FindRightIndex(int A[], int n, int target)
  {
    int mid;
    int low = 0, high = n - 1;
    while(low <= high)
    {
      mid = low + ((high - low) >> 1);
      if(A[mid] == target && (mid == (n - 1) || A[mid + 1] != target))
        return mid;
      if(A[mid] <= target)
        low = mid + 1;
      else 
        high = mid - 1;
    }
    return -1;
  }
};