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LeetCode-922. Sort Array By Parity II

2022-08-10 15:36:00 【51CTO】

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.
      
1.
2.
3.

Note:

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

题解：

       
       class 
       
       Solution {
       
       public:
       
       vector
       
       <
       
       int
       
       > 
       
       sortArrayByParityII(
       
       vector
       
       <
       
       int
       
       >& 
       
       A) {
       
       int 
       
       n 
       
       = 
       
       A.
       
       size();
       
       vector
       
       <
       
       int
       
       > 
       
       odd, 
       
       even;
       
       for (
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       n; 
       
       i
       
       ++) {
       
       if (
       
       A[
       
       i] 
       
       % 
       
       2 
       
       == 
       
       0) {
       
       even.
       
       push_back(
       
       A[
       
       i]);
       
      }
       
       else {
       
       odd.
       
       push_back(
       
       A[
       
       i]);
       
      }
       
    }
       
       A.
       
       clear();
       
       for (
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       n; 
       
       i
       
       ++) {
       
       if (
       
       i 
       
       % 
       
       2 
       
       == 
       
       0) {
       
       A.
       
       push_back(
       
       even.
       
       back());
       
       even.
       
       pop_back();
       
      }
       
       else {
       
       A.
       
       push_back(
       
       odd.
       
       back());
       
       odd.
       
       pop_back();
       
      }
       
    }
       
       return 
       
       A;
       
  }
       
};
      
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版权声明
本文为[51CTO]所创，转载请带上原文链接，感谢
https://blog.51cto.com/u_14150327/5564023

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