https://leetcode.com/problems/sort-array-by-parity-ii/
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
代码:
class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { int n = A.size(); vector<int> ans; vector<int> isodd; vector<int> nodd; for(int i = 0; i < n; i ++) { if(A[i] % 2) isodd.push_back(A[i]); else nodd.push_back(A[i]); } int a = isodd.size() - 1, b = nodd.size() - 1; for(int i = 0; i < n; i ++) { if(i % 2) {ans.push_back(isodd[a]); a --;} else {ans.push_back(nodd[b]); b --;} } return ans; } };
这个题好无聊哦。。。