Search for a Range

描述

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].

分析

已经排好了序，用二分查找。

重新实现 lower_bound 和 upper_bound

  1. // Search for a Range
  2. // 重新实现 lower_bound 和 upper_bound
  3. // 时间复杂度O(logn)，空间复杂度O(1)
  4. class Solution {
  5. public:
  6.     vector<int> searchRange (vector<int>& nums, int target) {
  7.         auto lower = lower_bound(nums.begin(), nums.end(), target);
  8.         auto uppper = upper_bound(lower, nums.end(), target);
  10.         if (lower == nums.end() || *lower != target)
  11.             return vector<int> { -1, -1 };
  12.         else
  13.             return vector<int> {distance(nums.begin(), lower), distance(nums.begin(), prev(uppper))};
  14.     }
  16.     template<typename ForwardIterator, typename T>
  17.     ForwardIterator lower_bound (ForwardIterator first,
  18.             ForwardIterator last, T value) {
  19.         while (first != last) {
  20.             auto mid = next(first, distance(first, last) / 2);
  22.             if (value > *mid) first = ++mid;
  23.             else              last = mid;
  24.         }
  26.         return first;
  27.     }
  29.     template<typename ForwardIterator, typename T>
  30.     ForwardIterator upper_bound (ForwardIterator first,
  31.             ForwardIterator last, T value) {
  32.         while (first != last) {
  33.             auto mid = next(first, distance (first, last) / 2);
  35.             if (value >= *mid) first = ++mid;  // 与 lower_bound 仅此不同
  36.             else               last = mid;
  37.         }
  39.         return first;
  40.     }
  41. };

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/search/search-for-a-range.html