33. Search in Rotated Sorted Array
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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.
第一种是直接顺序查找,时间复杂度O(N)
class Solution {public: int search(vector<int>& nums, int target) { int i; for(i=0;i<nums.size();i++) { if(target==nums[i]) return i; } return -1; }};
第二种方法是类似二分查找,先找到旋转后左边升序序列的最后一个最大数的位置pivot,如果target大于第一个数,则在0~pivot中二分查找target,否则在pivot+1到最后一个位置查找。
class Solution {public: int search(vector<int>& nums, int target) { int left=0,right=nums.size()-1,pivot,result; if(nums[left]<nums[right]) result=twodivide(nums,target,0,nums.size()-1); else { pivot=searchhelp(nums); if(target>=nums[left]) result=twodivide(nums,target,left,pivot); else result=twodivide(nums,target,pivot+1,right); } return result; } int twodivide(vector<int>& nums,int target,int left,int right) { int mid; while(left<=right) { mid=(left+right)/2; if(nums[mid]==target) return mid; else if(nums[mid]<target) left=mid+1; else right=mid-1; } return -1; } int searchhelp(vector<int>& nums) { int left=0; int right=nums.size()-1; while(left<right-1) { int mid=(left+right)/2; if(nums[mid]<nums[left]) right=mid; else left=mid; } return left; }};
第三种方法也是类似二分法,看到别人写的,主要分三种情况,写的很简洁易懂
class Solution {public: int search(vector<int>& nums, int target) { int l=0,r=nums.size()-1,mid; while(l<=r) { mid=(l+r)/2; if(target==nums[mid]) return mid; if(nums[l]<=nums[r]) { if(nums[mid]>target) r=mid-1; else l=mid+1; } else if(nums[l]<=nums[mid]) { if(target>nums[mid]||target<nums[l]) l=mid+1; else r=mid-1; } else { if(target<nums[mid]||target>=nums[l]) r=mid-1; else l=mid+1; } } return -1; }};
第四种方法:
循环递增数组有这么一个性质:以数组中间元素将循环递增数组划分为两部分,则一部分为一个严格递增数组,而另一部分为一个更小的循环递增数组。
当中间元素大于首元素时,前半部分为严格递增数组,后半部分为循环递增数组;当中间元素小于首元素时,前半部分为循环递增数组;后半部分为严格递增数组。
class Solution {public: int search(vector<int>& nums, int target) { int l=0,r=nums.size()-1,mid; while(l<=r) { mid=(l+r)/2; if(target==nums[mid]) return mid; else if(nums[l]<=nums[mid])//左边部分为有序数组 { if(target>=nums[l]&&target<nums[mid]) r=mid-1; else l=mid+1; } else//右边部分为有序数组 { if(target>nums[mid]&&target<=nums[r]) l=mid+1; else r=mid-1; } } return -1; }};
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