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. public class Solution {
  5.     public int[] searchRange(int[] nums, int target) {
  6.         int lower = lower_bound(nums, 0, nums.length, target);
  7.         int upper = upper_bound(nums, 0, nums.length, target);
  9.         if (lower == nums.length || nums[lower] != target)
  10.             return new int[]{-1, -1};
  11.         else
  12.             return new int[]{lower, upper-1};
  13.     }
  15.     int lower_bound (int[] A, int first, int last, int target) {
  16.         while (first != last) {
  17.             int mid = first + (last - first) / 2;
  18.             if (target > A[mid]) first = ++mid;
  19.             else                 last = mid;
  20.         }
  22.         return first;
  23.     }
  25.     int upper_bound (int[] A, int first, int last, int target) {
  26.         while (first != last) {
  27.             int mid = first + (last - first) / 2;
  28.             if (target >= A[mid]) first = ++mid;  // 与 lower_bound 仅此不同
  29.             else                  last = mid;
  30.         }
  32.         return first;
  33.     }
  34. }

相关题目

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