Leetcode 6. Searching for search in rotated Sorted array after an ordered array rotation

Search in rotated Sorted Array
Difficulty: Hard

Suppose a sorted array is rotated on 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 is given a target value to search. If found in the array is return its index, otherwise return-1.

Assume no duplicate exists in the array.

(The next topic will study the existence of duplicate cases search in rotated Sorted Array II)

Problem Solving Ideas:
For array search problems, there are roughly three time complexity O(N),O(Lg  N),O(1)
Where linear complexity o ( n ) , that is, sequential search, and constant time o ( 1 ) Generally implemented by keyword index or hash table, and logarithmic complexity < Span class= "Mrow" id= "mathjax-span-848" > o ( l g &NBSP, n ) , often by binary lookup, binary search tree type implementation

For the condition in the subject is the deformation of an ordered array, so you can think of "quick sort Selection" – Two-point lookup
the difficulty is to find the dividing element and the recursive condition and its boundary between the left and right sides of the dividing element .

For dividing elements, it is easy to think of directly taking m = (l+r) / 2; the intermediate array boundary [L, R]
Observe the example:
0 1 2 4 5 6 7
4 5 6 7 0 1 2
5 1 3
3 1
NUMS[M] > Nums[l]: (L, M-1) monocytogenes
NUMS[M] <= Nums[l]: (m+1, R) monocytogenes
Note:
-When the array takes the boundary [L, R], M takes to the center of the right (as in 2 elements, M = 1), so at this time, should compare nums[m], num[l]
-When the array boundary [L, R] is taken, M is centered on the left (as in 2 elements, M = 0), so at this time, it should be compared nums[m], num[r];

Array boundary is [L,r] – < Span class= "Mrow" id= "mathjax-span-42" > o ( l g &NBSP, n )

classSolution { Public://nums array boundary is [L,r]    intSEARCHR ( vector<int>& Nums,intLintRintTarget) {if(R <= L)return-1;intm = (l+r)/2;if(Nums[m] = = target)returnMif(Nums[l] < nums[m]) {if(Target >= nums[l] && target < nums[m])returnSEARCHR (Nums, L, M, target);Else                returnSEARCHR (Nums, m+1, R, Target); }Else{if(Target > Nums[m] && target <= nums[r-1])returnSEARCHR (Nums, m+1, R, Target);Else                returnSEARCHR (Nums, L, M, target); }    }intSearch vector<int>& Nums,intTarget) {returnSEARCHR (Nums,0, Nums.size (), target); }};

Array bounds [L, r]– Note the bounds of the incoming array

classSolution { Public://nums array boundary is [L,r]    intSEARCHR ( vector<int>& Nums,intLintRintTarget) {if(R < L)return-1;intm = (l+r)/2;if(Nums[m] = = target)returnMif(Nums[r] > Nums[m]) {if(Target > Nums[m] && target <= nums[r])returnSEARCHR (Nums, m+1, R, Target);Else                returnSEARCHR (Nums, L, M-1, target); }Else{if(Target >= nums[l] && target < nums[m])returnSEARCHR (Nums, L, M-1, target);Else                returnSEARCHR (Nums, m+1, R, Target); }    }intSearch vector<int>& Nums,intTarget) {returnSEARCHR (Nums,0, Nums.size ()-1, target); }};

Sequential search O(n) – Not recommended

class  solution {public : int  search (vector  <int ;  & Nums, int  target" {int  i = 0 ; for  (; i < nums.size (); i++) {if  (nums[i] = = target) break;  } if  (nums.size () = = i) return -1 ;            else     return  i; }};

(The next topic will study the existence of duplicate cases search in rotated Sorted Array II)

Leetcode 6. Searching for search in rotated Sorted array after an ordered array rotation