[LeetCode] 1470. Shuffle the Array c+

• 문제

Given the array nums consisting of 2n elements in the form [x1,x2,...,xn,y1,y2,...,yn].

Return the array in the form [x1,y1,x2,y2,...,xn,yn].

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7] 
Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].

 

• 문제 해결

class Solution {
public:
    vector<int> shuffle(vector<int>& nums, int n) {
        vector<int> answer;
        for(int i = 0; i < n; i++){
            answer.push_back(nums[i]);
            answer.push_back(nums[i+n]);
        }
        return answer;
    }
};

다른 방법

class Solution {
public:
    vector<int> shuffle(vector<int>& nums, int n) {
        
        int len = nums.size();
        
		// to store the pair of numbers in right half of the original array
        for(int i = n; i < nums.size(); i++) {
            nums[i] = (nums[i] * 1024) + nums[i - n];
        }
        
        int index = 0;
		// to retrive values from the pair of numbers and placing those retrieved value at their desired position
        for(int i = n; i < nums.size(); i++, index += 2) {
            nums[index] = nums[i] % 1024;
            nums[index + 1] = nums[i] / 1024;
        }
        
        return nums;
        
    }
};

 

다른 방법2

class Solution {
public:
    vector<int> shuffle(vector<int>& nums, int n) {
        
        int len = nums.size();
        
        for(int i = n; i < len; i++) {
            nums[i] = (nums[i] << 10) | nums[i - n];
        }
        
        int index = 0;
        for(int i = n; i < len; i++, index += 2) {
            nums[index] = nums[i] & 1023;
            nums[index + 1] = nums[i] >> 10;
        }
        
        return nums;
        
    }
};

https://leetcode.com/problems/shuffle-the-array/discuss/1314638/JAVA-C%2B%2B-%3A-Simple-or-O(1)-Space-In-Place-or-100-Faster-or-Efficient-or-Explained 

✅ [JAVA / C++] : Simple | O(1) Space In-Place | 100% Faster | Efficient | Explained - LeetCode Discuss