LeetCode81——Search in Rotated Sorted Array II
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LeetCode81——Search in Rotated Sorted Array II
加入重复数据的Search in Rotated Sorted Array
原题
Follow up for “Search in Rotated Sorted Array”:
What if duplicates are allowed?
在Search in Rotated Sorted Array题中,如果允许重复元素情况求解。
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
无重复元素
进行二分查找,以中间元素nums[mid]
为中心分为左右两边,必然出现下述情况的某一种
左边有序
右边有序
以左边有序为例,假如左边有序,那么我们判断target
是否在左边子序列中,如果在,则对子序列进行二分搜索,否则对右边子序列做二分搜索。
代码(无重复元素Roated Sorted Array的二分查找):
class Solution {public: int search(vector<int>& nums, int target) { int nSize = nums.size(); int low = 0; int high = nSize - 1; while (low <= high) { int mid = low + (high - low) / 2;//防止溢出 if (target == nums[mid])//查找结果成功 return mid; //这里的细节要用>=原因是计算mid的时候是向下取整有可能 //mid==low那么nums[mid]==nums[low] //排除测试用例 [3,1] 1 else if (nums[mid] >= nums[low])//左边有序 { if (target < nums[mid] && target>=nums[low])//target在左边 high = mid - 1; else//target在右边 low = mid + 1; } else//右边有序 { if (target > nums[mid] && target <= nums[high]) low = mid + 1; else high = mid - 1; } } return -1; }};
有重复元素
那么现在问题来了,如何去除重复元素呢?
其实考虑的的还是一样,只不过比较的时候考虑一点,要考虑当nums[mid]==nums[low]时的情况,也就是如果给定序列为[6,6,6,6,6,4,5],而查找的target是5时,我们在第一次循环中,得到的nums[mid]==nums[low]这个时候我们要舍去这个nums[low],使low++,可以理解为整体右移(由于mid是向下取整,所以最坏情况下需要至少右移两次),这样就去掉重复部分了。
代码:
class Solution {public: bool search(vector<int>& nums, int target) { int nSize = nums.size(); int low = 0; int high = nSize - 1; while (low <= high) { int mid = low + (high - low) / 2; if (nums[mid] == target) return true; else if (nums[low] < nums[mid])//右边有序 { if (target >= nums[low] && target<nums[mid]) high = mid - 1; else low = mid + 1; } else if (nums[low] > nums[mid])//左边有序 { if (target <= nums[high] && target > nums[mid]) low = mid + 1; else high = mid - 1; } else//相等要"整体右移" { low++; } } return false; }};
参考:
http://blog.csdn.net/linhuanmars/article/details/20588511
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