Binary search: Difference between revisions
Content added Content deleted
m (Forgot the task tag) |
|||
| Line 33: | Line 33: | ||
==[[Java]]== |
==[[Java]]== |
||
[[Category:Java]] |
[[Category:Java]] |
||
==Iterative== |
|||
... |
|||
//check will be the number we are looking for |
|||
//nums will be the array we are searching through |
|||
int hi = nums.length - 1; |
|||
int lo = 0; |
|||
int guess = (hi + lo) / 2; |
|||
while((nums[guess] != check) && (hi > lo)){ |
|||
if(nums[guess] > check){ |
|||
hi = guess - 1; |
|||
}else if(nums[guess] < check){ |
|||
lo = guess + 1; |
|||
} |
|||
guess = (hi + lo) / 2; |
|||
} |
|||
if(hi < lo){ |
|||
System.out.println(check + " not in array"); |
|||
}else{ |
|||
System.out.println("found " + nums[guess] + " at index " + guess); |
|||
} |
|||
... |
|||
Revision as of 06:47, 8 November 2007
You are encouraged to solve this task according to the task description, using any language you may know.
Do a binary search through a sorted integer array for a certain number. Implementations can be recursive or iterative (both if you can). Print out whether or not the number was in the array afterwards. If it was, print the index also. The algorithms are as such (from the wikipedia):
Recursive:
BinarySearch(A[0..N-1], value, low, high) {
if (high < low)
return not_found
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch(A, value, low, mid-1)
else if (A[mid] < value)
return BinarySearch(A, value, mid+1, high)
else
return mid
}
Iterative:
BinarySearch(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else if (A[mid] < value)
low = mid + 1
else
return mid
}
return not_found
}