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.
分析
如果允许重复元素，则Search in Rotated Sorted Array题中如果 A[m]>=A[l], 那么 [l,m] 为递增序列的假设就不能成立了，比
如 [1,2，3,1,1,1]。
如果 A[m]>=A[l] 不能确定递增，那就把它拆分成两个条件：
• 若 A[m]>A[l]，则区间 [l,m] 一定递增
• 若 A[m]==A[l] 确定不了，那就 l++，往下看一步即可。
Search in Rotated Sorted Array 的code：
1 class Solution { 2 public: 3 bool search(int A[], int n, int target) { 4 int low = 0; 5 int high = n-1; 6 int mid ; 7 8 9 while(low<=high) 10 { 11 mid= (low + high)/2; 12 13 if(A[mid]==target) 14 { 15 return true; 16 } 17 18 if(A[low]<=A[mid]) // the pivot is in the bottom half 19 { 20 if(A[low]<=target && target < A[mid]) //target in the first half 21 high = mid-1; 22 else 23 low = mid+1;//target in the bottom half 24 } 25 else // the pivot is in the first half 26 { 27 if(A[mid] < target && target <= A[high])// the pivot is in the first half 28 low = mid+1; 29 else //target in the first half 30 high = mid-1; 31 } 32 33 } 34 return false; 35 36 } 37 };
将if (A[low] <= A[mid]) 拆成if (A[low] < A[mid]) 和 if (A[low] == A[mid]) 两个分支，code如下：
1 class Solution { 2 public: 3 bool search(int A[], int n, int target) { 4 int low = 0; 5 int high = n-1; 6 int mid ; 7 8 9 while(low<=high) 10 { 11 mid= (low + high)/2; 12 13 if(A[mid]==target) 14 { 15 return true; 16 } 17 18 if(A[low]<A[mid]) 19 { 20 if(A[low]<=target && target < A[mid]) 21 high = mid-1; 22 else 23 low = mid+1; 24 } 25 else if(A[low] > A[mid]) 26 { 27 if(A[mid] < target && target <= A[high]) 28 low = mid+1; 29 else 30 high = mid-1; 31 } 32 else 33 low++; 34 35 } 36 return false; 37 38 } 39 };