tags: array, binary search, leetcode
问题描述
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
和上一个问题的唯一不同点是，数组里面的元素可能是重复的。
为此需要剔除重复的元素。1
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55int find(int a[], int n)
{
if(n==0) return -1;
if(n==1) return 0;
if(n==2) return a[0]<=a[1]?0:1;
int left=0,right=n-1, mid;
while(right-left>1)
{
mid = left + (right-left)/2;
if(a[left] < a[mid])
left = mid;
else if(a[left] > a[mid])
right = mid;
else left++;// for case 2,2,2,0,2,2 a[left] == a[mid],
//we just ignore the leftmost one
}
if(a[left]>a[right])
return right;
else
return 0;
}
int binarySearch(int a[], int left, int right, int target)
{
int mid;
while(left<=right)
{
mid = left + (right-left)/2;
if(a[mid] == target)
return mid;
else if(a[mid] < target)
left = mid+1;
else
right = mid -1;
}
return -1;
}
bool search(int A[], int n, int target) {
//three steps: cut the array into two sorted arrays;
// search the target in tow sorted arrays;
// return the index;
int index = find(A,n);
int result = false;
if(index == 0)
result = binarySearch(A,0,n-1,target);
else
{
if(target >= A[0])
result = binarySearch(A,0,index-1,target);
else
result = binarySearch(A,index,n-1,target);
}
if(result != -1) return true;
else return false;
}