37 lines
1.4 KiB
C++
37 lines
1.4 KiB
C++
#include <vector>
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#include <algorithm>
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/**
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* 81. Search in Rotated Sorted Array II
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* There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).
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* Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].
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* Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.
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* You must decrease the overall operation steps as much as possible.
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*
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* Analysis:
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* class Solution has a complexity of O(n).
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* class SolutionBinary has O(n + 2 * log(n)) -> O(n).
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*
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* But which is faster? I sorely have no idea.
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*/
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class Solution {
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public:
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static bool search(const std::vector<int>& nums, int target) {
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return std::any_of(nums.begin(), nums.end(), [&](int x){ return x == target; });
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}
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};
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class SolutionBinary {
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public:
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static bool search(const std::vector<int>& nums, int target) {
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auto it = std::next(nums.begin());
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for (; it != nums.end(); ++it)
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if (*it < *std::prev(it))
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break;
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if (it == nums.end())
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return std::binary_search(nums.begin(), nums.end(), target);
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return std::binary_search(nums.begin(), it, target) || std::binary_search(it, nums.end(), target);
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}
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};
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