136. Single Number
Given a non-empty array of integers, every element appears twice
except for one. Find that single one.Note:
Your algorithm should have a linear runtime complexity. Could you
implement it without using extra memory?Example 1:
Input: [2,2,1] Output: 1
Example 2:
Input: [4,1,2,1,2] Output: 4
My O(n) solution using XOR：
class Solution {
int singleNumber(int A[], int n) {
int result = 0;
for (int i = 0; i<n; i++)
{
result ^=A[i];
}
return result;
}
}
Easiest way to solve by using bit manipulation：
Logic: XOR will return 1 only on two different bits. So if two numbers
are the same, XOR will return 0. Finally only one number left. A ^ A =
0 and A ^ B ^ A = B.
class Solution {
public:
int singleNumber(int A[], int n) {
int result=A[0];
for(int i=1;i<n;i++)
{
result= result^A[i]; /* Get the xor of all elements */
}
return result;
}
};
169. Majority Element
Given an array of size n, find the majority element. The majority
element is the element that appears more than ⌊ n/2 ⌋ times.You may assume that the array is non-empty and the majority element
always exist in the array.Example 1:
Input: [3,2,3] Output: 3
Example 2:
Input: [2,2,1,1,1,2,2] Output: 2
O(n) time O(1) space fastest solution:
public class Solution {
public int majorityElement(int[] num) {
int major=num[0], count = 1;
for(int i=1; i<num.length;i++){
if(count==0){
count++;
major=num[i];
}else if(major==num[i]){
count++;
}else count--;
}
return major;
}
}
c++:
Hash Table:
The hash-table solution is very straightforward. We maintain a mapping
from each element to its number of appearances. While constructing the
mapping, we update the majority element based on the max number of
appearances we have seen. Notice that we do not need to construct the
full mapping when we see that an element has appeared more than n / 2
times.The code is as follows, which should be self-explanatory.
class Solution {
public:
int majorityElement(vector<int>& nums) {
unordered_map<int, int> counts;
int n = nums.size();
for (int i = 0; i < n; i++)
if (++counts[nums[i]] > n / 2)
return nums[i];
}
};
Sorting
Since the majority element appears more than n / 2 times, the n / 2-th
element in the sorted nums must be the majority element. This can be
proved intuitively. Note that the majority element will take more than
n / 2 positions in the sorted nums (cover more than half of nums). If
the first of it appears in the 0-th position, it will also appear in
the n / 2-th position to cover more than half of nums. It is similar
if the last of it appears in the n - 1-th position. These two cases
are that the contiguous chunk of the majority element is to the
leftmost and the rightmost in nums. For other cases (imagine the chunk
moves between the left and the right end), it must also appear in the
n / 2-th position.The code is as follows, being very short if we use the system
nth_element (thanks for @qeatzy for pointing out such a nice
function).
class Solution {
public:
int majorityElement(vector<int>& nums) {
nth_element(nums.begin(), nums.begin() + nums.size() / 2, nums.end());
return nums[nums.size() / 2];
}
};
Randomization
This is a really nice idea and works pretty well (16ms running time on
the OJ, almost fastest among the C++ solutions). The proof is already
given in the suggested solutions.The code is as follows, randomly pick an element and see if it is the
majority one.
class Solution {
public:
int majorityElement(vector<int>& nums) {
int n = nums.size();
srand(unsigned(time(NULL)));
while (true) {
int idx = rand() % n;
int candidate = nums[idx];
int counts = 0;
for (int i = 0; i < n; i++)
if (nums[i] == candidate)
counts++;
if (counts > n / 2) return candidate;
}
}
};
Divide and Conquer
This idea is very algorithmic. However, the implementation of it
requires some careful thought about the base cases of the recursion.
The base case is that when the array has only one element, then it is
the majority one. This solution takes 24ms.
class Solution {
public:
int majorityElement(vector<int>& nums) {
return majority(nums, 0, nums.size() - 1);
}
private:
int majority(vector<int>& nums, int left, int right) {
if (left == right) return nums[left];
int mid = left + ((right - left) >> 1);
int lm = majority(nums, left, mid);
int rm = majority(nums, mid + 1, right);
if (lm == rm) return lm;
return count(nums.begin() + left, nums.begin() + right + 1, lm) > count(nums.begin() + left, nums.begin() + right + 1, rm) ? lm : rm;
}
};
Moore Voting Algorithm
A brilliant and easy-to-implement algorithm! It also runs very fast,
about 20ms.
class Solution {
public:
int majorityElement(vector<int>& nums) {
int major, counts = 0, n = nums.size();
for (int i = 0; i < n; i++) {
if (!counts) {
major = nums[i];
counts = 1;
}
else counts += (nums[i] == major) ? 1 : -1;
}
return major;
}
};
Bit Manipulation
Another nice idea! The key lies in how to count the number of 1’s on a
specific bit. Specifically, you need a mask with a 1 on the i-the bit
and 0 otherwise to get the i-th bit of each element in nums. The code
is as follows.
class Solution {
public:
int majorityElement(vector<int>& nums) {
int major = 0, n = nums.size();
for (int i = 0, mask = 1; i < 32; i++, mask <<= 1) {
int bitCounts = 0;
for (int j = 0; j < n; j++) {
if (nums[j] & mask) bitCounts++;
if (bitCounts > n / 2) {
major |= mask;
break;
}
}
}
return major;
}
};
461. Hamming Distance
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given two integers x and y, calculate the Hamming distance.
Java 1 Line Solution ：
What does come to your mind first when you see this sentence
“corresponding bits are different”? Yes, XOR! Also, do not forget there is a decent function Java provided: Integer.bitCount() ~~~
class Solution {
public class Solution {
public int hammingDistance(int x, int y) {
return Integer.bitCount(x ^ y);
}
}
}
Java 3-Line Solution：
class Solution {
public int hammingDistance(int x, int y) {
int xor = x ^ y, count = 0;
for (int i=0;i<32;i++) count += (xor >> i) & 1;
return count;
}
}