LintCode (63) Search in Rotated Array II

Search in Rotated Sorted Array II

40%

Accepted

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

和find min in rotated array with dup是一样的，如果有dup，判断左旋和右旋无法判断的时候，移动beg,这样，最差复杂度为o(n)

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class Solution { /** * param A : an integer ratated sorted array and duplicates are allowed * param target : an integer to be search * return : a boolean */ public: bool search(vector<int> &A, int target) { // write your code here if (A.empty()) return false; int beg=0; int end=A.size()-1; while(beg+1<end){ int mid=beg+(end-beg)/2; if (A[mid]==target) return true; if (A[mid]>A[beg]){ if (A[mid]>=target && target>=A[beg]){ end=mid-1; } else{ beg=mid+1; } } else if (A[mid]<A[beg]){ if (A[mid]<=target && target<=A[end]){ beg=mid+1; } else{ end=mid-1; } } else{ beg++; } } return (A[beg]==target || A[end]==target); } };