Binary Search — Classic Exact Target
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Problem 301 · Binary Search
Difficulty: Easy · Pattern: Classic Exact Target
Solutions
// C
int search(int* nums, int n, int target) {
int lo=0, hi=n-1;
while (lo<=hi) {
int mid=lo+(hi-lo)/2;
if (nums[mid]==target) return mid;
else if (nums[mid]<target) lo=mid+1;
else hi=mid-1;
}
return -1;
}
// C++
int search(vector<int>& nums, int target) {
int lo=0, hi=nums.size()-1;
while (lo<=hi) {
int mid=lo+(hi-lo)/2;
if (nums[mid]==target) return mid;
else if (nums[mid]<target) lo=mid+1;
else hi=mid-1;
}
return -1;
}
// Java
public int search(int[] nums, int target) {
int lo=0, hi=nums.length-1;
while (lo<=hi) {
int mid=lo+(hi-lo)/2;
if (nums[mid]==target) return mid;
else if (nums[mid]<target) lo=mid+1;
else hi=mid-1;
}
return -1;
}
# Python
def search(nums, target):
lo, hi = 0, len(nums)-1
while lo <= hi:
mid = lo + (hi-lo)//2
if nums[mid] == target: return mid
elif nums[mid] < target: lo = mid+1
else: hi = mid-1
return -1
Complexity
- Time: O(log n)
- Space: O(1)
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