二分查找是我們經常會遇到的算法,思路清晰,代碼簡潔。二分查找要求序列有序,且支持隨機存取,一般情況下我們討論的序列不存在相同元素,則二分查找可以很熟練的表示如下:
int binsearch(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
return mid;
}
}
return -1;
}
但是當我們對有序序列不做要求時,即可能出現相同的元素的情況下,二分查找就回出現一些擴展的問題,比如:
- 返回key的第一次出現的下標,若沒有返回-1;
- 返回key最後一次出現的下標,若沒有返回-1;
- 返回剛好小於key的元素的下標,若沒有返回-1;
- 返回剛好大於key的元素的下標,若沒有返回-1;
int searchLowerBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
right = mid-1;
}
}
return res;
}
對於(2),同理可得:int searchHigherBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
left = mid+1;
}
}
return res;
}
對於(3),使得key=key-1,則轉換成問題(2),對於(4),使得key=key+1,則轉換成問題(1)。Given a sorted array of integers, find the starting and ending position of a given target value.Your algorithm's runtime complexity must be in the order of O(log n).If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
這個題就是要求出最小的value的下標和最大的value的下標,所以代碼易得:
class Solution {
public:
int searchLowerBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
right = mid-1;
}
}
return res;
}
int searchHigherBound(int A[], int n, int target)
{
int left=0,right=n-1,res=-1;
while(left<=right)
{
int mid = left+((right-left)>>1);
if(A[mid]<target)
left = mid+1;
else if(A[mid]>target)
right = mid-1;
else
{
res = mid;
left = mid+1;
}
}
return res;
}
vector<int> searchRange(int A[], int n, int target) {
vector<int> res;
res.push_back(searchLowerBound(A,n,target));
res.push_back(searchHigherBound(A,n,target));
return res;
}
};