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 half19 {20 if(A[low]<=target && target < A[mid]) //target in the first half21 high = mid-1;22 else23 low = mid+1;//target in the bottom half24 }25 else // the pivot is in the first half26 {27 if(A[mid] < target && target <= A[high])// the pivot is in the first half28 low = mid+1;29 else //target in the first half30 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])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 };