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隔离偶数和奇数，保持顺序，O（1）空间，O（N）时间复杂度

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Input  = {12, 34, 45, 9, 8, 90, 3} 
Output = {12, 34, 8, 90, 45, 9, 3}

给定一个整数数组，在所有奇数之前重新排列所有偶数整数，但使用O（1）空间和O（n）时间复杂度将它们的原始序列保持在数组中 .

思想：

Algorithm: segregateEvenOdd()
1) Initialize two index variables left and right:  
            left = 0,  right = size -1 
2) Keep incrementing left index until we see an odd number.
3) Keep decrementing right index until we see an even number.
4) If lef < right then swap arr[left] and arr[right]

但这不能保证订单是一样的 .

如果我想使用O（1）空间来解决这个问题怎么办？

algorithm sorting

1 回答

  • -1

    你必须存储最后的奇数和最后的偶数位置，

    1. Initialize both last_odd, last_even to -1;
2. Iterate over array
   - Check if array[i] is odd or even,
   - if diff(last_odd/last_even,i) >= 2 and last_odd/even not equal to 
     -1:
        if (element is odd/even) new index = last_odd/even+1
        - store element value in temp
        - move elements from new_index up to i one to right 
          starting back from i-1 down to new_index.
        - store temp in new_index
        - store last_odd/even as new_index accordingly and add to 
          last_even/odd the diff(which is +1)
   - else store last_odd/even as i accordingly
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