leetcode_2 Sort Array By Parity II(sort)

• 题目描述：

  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.

• 中文描述：将给定的非负数组进行排序重组，使得下标为奇数，则下标对应的值为奇数，下标为偶数，则对应的值为偶数
• 解题思路：
  • 思路一：首先对数组遍历一次分别将偶数奇数存放，然后重新遍历数组依次从偶数数组，奇数数组中拿出值为数组赋值，时间复杂度为o(n)空间为o(n)
  • 思路二：将数组按照下标的奇偶性进行遍历，分别遇到第一个偶数下标不为偶数值，奇数下标，不为奇数值，将其互换，时间复杂度o(n)空间o(1)
• 解题代码：
  • 思路一的c++实现

    class Solution {
    public:
        vector<int> sortArrayByParityII(vector<int>& A) {
            vector<int> even;
            vector<int> odd;
            vector<int>::iterator it;
            for(it=A.begin();it!=A.end();it++)
            {
                if(*it%2==0)
                {
                    even.push_back(*it);
                }
                else{
                    odd.push_back(*it);
                }
            }
            int length=A.size();
            int i=0,j=0;
            int cy=0;
            bool flag=true;
            while(cy<length)
            {
                if(flag)
                {
                    A[cy++]=even[i++];
                    flag=false;
                }
                else{
                    A[cy++]=odd[j++];
                    flag=true;
                }
            }
            return A;
            
        }
    };
    

  • 思路二的python3实现

    class Solution:
        def sortArrayByParityII(self, A: List[int]) -> List[int]:
            length=len(A)
            ev_cy,od_cy=0,1
            while ev_cy<length and od_cy<length:
                if A[ev_cy]%2==0:
                    ev_cy+=2
                elif A[od_cy]%2==1:
                    od_cy+=2
                else:
                    A[ev_cy],A[od_cy]=A[od_cy],A[ev_cy]
                    ev_cy+=2
                    od_cy+=2
            return A
    

• 语法总结：
  • c++:迭代器声名语句：vector<int> iterator; vector.push_back() ;bool flag=true;vector.pop() 删除栈顶元素，无返回值。
  • python:flag=True;list 存在属性 list.pop(i)弹出某一元素，返回值为弹出元素值，默认值为末尾元素。