Search in Rotated Sorted Array II [LeetCode]
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.
Summary: Almost the same to Search in Rotated Sorted Array without duplicates, but careful about the corner case like [1311], we should search both left half and right half. So, in this case, the run-time complexity will between O(logN) to O(N).
1 class Solution { 2 public: 3 bool search(int A[], int n, int target) { 4 if(n == 0) 5 return false; 6 if(n == 1) { 7 if(A[0] == target ) 8 return true; 9 else 10 return false; 11 } 12 13 int median = n/2; 14 if(A[median] == target) 15 return true; 16 if(A[0] == target) 17 return true; 18 19 if(A[0] < A[median] && target > A[0] && target < A[median] || 20 (A[0] > A[median] && (target > A[0] || target < A[median]))){ 21 return search(A + 1, median - 0, target); 22 }else if(A[0] == A[median]){ 23 bool ret1 = search(A + 1, median - 0, target); 24 bool ret2 = false; 25 if(n - median - 1 <= 0) 26 ret2 = false; 27 28 ret2 = search(A + median + 1, n - median -1, target); 29 return ret1 || ret2; 30 }else{ 31 if(n - median - 1 <= 0) 32 return false; 33 34 return search(A + median + 1, n - median -1, target); 35 } 36 } 37 };
