922. Sort Array By Parity II

2018/10/14 21:29
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Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

 

Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

 

Note:

  1. 2 <= A.length <= 20000
  2. A.length % 2 == 0
  3. 0 <= A[i] <= 1000

题目说有一半是奇数,一半是偶数。这是一个简化。降低了处理边界的难度。

#include<vector>
#include <cstdlib>
#include<iostream>
#include <unordered_set>
#include <algorithm>
#include<string>
#include<queue>

using namespace std;

//Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.
//
//Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.
//
//You may return any answer array that satisfies this condition.
//
//
//
//Example 1:
//
//Input: [4,2,5,7]
//Output: [4,5,2,7]
//Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
//Note:
//
//2 <= A.length <= 20000
//A.length % 2 == 0
//0 <= A[i] <= 1000

class Solution {
public:
    vector<int> sortArrayByParityII(vector<int> &A) {
        int i = -1;
        int n = A.size();
        for (int j = 0; j < n; j = j + 2) {
            if (A[j] % 2 == 0) {
                continue;
            } else {
                //A[j]%2==1 at this moment
                i += 2;
                while (A[i] % 2 == 1)
                    i += 2;
                //A[i]%2==0 at this monent
                //cout << "i " << i << " j " << j << endl;
                swap(A[i], A[j]);
            }
        }
        return A;
    }
};

int main() {

    Solution solution;
    vector<int> A{4, 2, 5, 7};
    vector<int> res = solution.sortArrayByParityII(A);
    for (auto a:res)
        cout << a << " ";
    cout << endl;
    return 0;
}

 

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