922. Sort Array By Parity II (Easy)

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:

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

 

Solutions

class Solution {
    private void swap(int[] A, int i, int j) {
        int tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
    }

    public int[] sortArrayByParityII(int[] A) {
        int size = A.length;

        // stands for the even index pointer
        int evenPtr = 0;

        // stands for the odd index pointer
        int oddPtr = 1;

        for (int i = 0; i < size; i++) {
            // find odd value at even index
            while (evenPtr < size) {
                if (A[evenPtr] % 2 != 0) {
                    break;
                }

                evenPtr += 2;
            }

            // find even value at odd index
            while (oddPtr < size) {
                if (A[oddPtr] % 2 != 1) {
                    break;
                }

                oddPtr += 2;
            }

            if (evenPtr < size && oddPtr < size) {
                swap(A, evenPtr, oddPtr);
            }
        }

        return A;
    }
}

Incorrect Solutions

References

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