81.Search in Rotated Sorted Array II
2017-03-16 23:29
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81.Search in Rotated Sorted Array II
2017-03-16
本作品采用“Creative Commons 署名-非商业性使用-相同方式共享 3.0 Unported
(cc by-nc-sa)”进行许可。http://creativecommons.org/licenses/by-nc-sa/3.0/
所有代码参考:LeetCode题解,151道题完整版
因为允许重复,会出现first位和mid位数字虽然相等、但这个区间并非递增的情况,所以要单独拿出来判断一下
-FIN-
2017-03-16
本作品采用“Creative Commons 署名-非商业性使用-相同方式共享 3.0 Unported
(cc by-nc-sa)”进行许可。http://creativecommons.org/licenses/by-nc-sa/3.0/
所有代码参考:LeetCode题解,151道题完整版
version 1:
33.Search in Rotated Sorted Array的进阶因为允许重复,会出现first位和mid位数字虽然相等、但这个区间并非递增的情况,所以要单独拿出来判断一下
class Solution { public: bool search(vector<int>& nums, int target) { int first = 0, last = nums.size(); while(first != last) { int mid = first + (last - first) / 2; if(nums[mid] == target) return true; if(nums[first] < nums[mid]) { if(nums[first] <= target && target < nums[mid]) last = mid; else first = mid+1; } else if(nums[first] > nums[mid]) { if(nums[mid] < target && target <= nums[last-1]) first = mid+1; else last = mid; } else//nums[first] == nums[mid] { first++; } } return false; } };
分析:
时间复杂度是多少呢?最坏的情况下,应该是O(n)的, 形如:1 1 1 1 1 1 3(个人看法,欢迎指正)-FIN-
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