Binary search
Given an n-element ordered (ascending) integer array nums and a target value target, write a function to search the target in nums. If the target value exists, return the subscript, otherwise return - 1.
Example 1:
Input: num = [- 1,0,3,5,9,12], target = 9, output: 4
Explanation: 9 appears in nums and the subscript is 4
Example 2:
Input: num = [- 1,0,3,5,9,12], target = 2, output: - 1
Explanation: 2 does not exist in nums, so - 1 is returned
# python version
class Solution:
def search(self, nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
middle = (left + right) // 2
if nums[middle] < target:
left = middle + 1
elif nums[middle] > target:
right = middle - 1
else:
return middle
return -1
# java version
class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
while(left <= right){
int mid = (left+right)/2;
if(nums[mid]<target){
left = mid + 1;
}else if(nums[mid]>target){
right = mid - 1;
}else{
return mid;
}
}
return -1;
}
}
Given a sort array and a target value, find the target value in the array and return its index. If the target value does not exist in the array, returns the position where it will be inserted in order.
You must use an algorithm with a time complexity of O(log n).
Example 1:
Input: num = [1,3,5,6], target = 5, output: 2
Example 2:
Input: num = [1,3,5,6], target = 2, output: 1
# python
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
n = len(nums)
left = 0
right = n - 1
while(left<=right):
mid = (left + right)//2
if nums[mid]>target:
right = mid - 1
elif nums[mid]<target:
left = mid + 1
else:
# 1. The target value is equal to an element in the array return mid;
return mid
# 2. The target value is before all elements of the array. 3. The target value is inserted into the array. 4. The target value is after all elements of the array. return right + 1;
return right+1
# java
class Solution {
public int searchInsert(int[] nums, int target) {
int n = nums.length;
int left = 0;
int right = n-1;
while(left<=right){
int mid = (left+right)/2;
if(nums[mid]<target){
left = mid + 1;
}else if(nums[mid]>target){
right = mid - 1;
}else{
return mid;
}
}
return right + 1;
}
}
Given an integer array nums arranged in ascending order and a target value target. Find the start and end positions of the given target value in the array.
If the target value target does not exist in the array, return [- 1, - 1].
Advanced:
Can you design and implement an algorithm with time complexity of O(log n) to solve this problem?
Example 1:
Input: num = [5,7,7,8,8,10], target = 8, output: [3,4]
Example 2:
Input: num = [5,7,7,8,8,10], target = 6, output: [- 1, - 1]
# python
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
# Binary search to find the left boundary of target (excluding target)
def getLeftBorder(nums,target):
n = len(nums)
left = 0
right = n - 1
leftBorder = -2
while(left <= right):
mid = (left + right)//2
if nums[mid] >= target:
right = mid - 1
leftBorder = right
else: # nums[mid] < target
left = mid + 1
return leftBorder
# Binary search to find the right boundary of target (excluding target)
def getRightBorder(nums,target):
n = len(nums)
left = 0
right = n - 1
rightBorder = -2
while(left <= right):
mid = (left + right)//2
if nums[mid] <= target:
left = mid + 1
rightBorder = left
else: # nums[mid] > target
right = mid - 1
return rightBorder
leftBorder = getLeftBorder(nums,target)
rightBorder = getRightBorder(nums,target)
if(leftBorder == -2 or rightBorder == -2):
return [-1,-1]
elif((rightBorder-leftBorder)>1):
return [leftBorder+1,rightBorder-1]
else:
return [-1,-1]
# java
class Solution {
public int[] searchRange(int[] nums, int target) {
int leftBorder = getLeftBorder(nums, target);
int rightBorder = getRightBorder(nums, target);
if(leftBorder==-2||rightBorder==-2){
return new int[]{-1,-1};
}else if(rightBorder-leftBorder>1){
return new int[]{leftBorder+1,rightBorder-1};
}else{
return new int[]{-1,-1};
}
}
public int getLeftBorder(int[] nums, int target){
int n = nums.length;
int left = 0;
int right = n - 1;
int leftBorder = -2;
while(left<=right){
int mid = (left+right)/2;
if(nums[mid]<target){
left = mid + 1;
}else{ //nums[mid]>=target
right = mid - 1;
leftBorder = right;
}
}
return leftBorder;
}
public int getRightBorder(int[] nums, int target){
int n = nums.length;
int left = 0;
int right = n - 1;
int rightBorder = -2;
while(left<=right){
int mid = (left+right)/2;
if(nums[mid]>target){
right = mid - 1;
}else{ //nums[mid]<=target
left = mid + 1;
rightBorder = left;
}
}
return rightBorder;
}
}
Give you a nonnegative integer x, calculate and return the square root of X.
Since the return type is an integer, only the integer part will be retained and the decimal part will be rounded off.
Note: it is not allowed to use any built-in exponential functions and operators, such as pow(x, 0.5) or x ** 0.5.
Example 1:
Input: x = 4 Output: 2
Example 2:
Input: x = 8 output: 2
Explanation: the square root of 8 is 2.82842... Because the return type is an integer, the decimal part will be rounded off.
# python
class Solution:
def mySqrt(self, x: int) -> int:
left = 0
right = x
while(left<=right):
mid = (left + right)//2
if(mid*mid>x):
right = mid - 1
elif(mid*mid<x):
left = mid + 1
else:
return mid
return right
# java
class Solution {
public int mySqrt(int x) {
int left = 0;
int right = x;
while(left<=right){
int mid = (left+right)/2;
if((long)mid*mid<x){
left = mid + 1;
}else if((long)mid*mid>x){
right = mid - 1;
}else{
return mid;
}
}
return right;
}
}
Given a positive integer num, write a function. If num is a complete square number, it returns true, otherwise it returns false.
Advanced: do not use any built-in library functions, such as sqrt.
Example 1:
Input: num = 16 output: true
Example 2:
Input: num = 14 output: false
# python
class Solution:
def isPerfectSquare(self, num: int) -> bool:
left = 0
right = num
while(left<=right):
mid = (left+right)//2
if(mid*mid<num):
left = mid + 1
elif(mid*mid>num):
right = mid - 1
else:
return True
return False
# java
class Solution {
public boolean isPerfectSquare(int num) {
int left = 0;
int right = num;
while(left<=right){
int mid = (left+right)/2;
if((long)mid*mid<num){
left = mid + 1;
}else if((long)mid*mid>num){
right = mid - 1;
}else{
return true;
}
}
return false;
}
}