LeetCode之Sort Array By Parity II（Kotlin）

問題： 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.
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方法： 兩個index同時遍歷，若是遇到兩個數組元素都不符合規則則互相交換位置，遍歷整個數組後輸出數組則爲最終結果。

具體實現：

class SortArrayByParityII {
    fun sortArrayByParityII(A: IntArray): IntArray {
        var oddIndex = 1
        var evenIndex = 0
        var temp : Int
        while(oddIndex <= A.lastIndex && evenIndex <= A.lastIndex) {
            if (A[evenIndex] % 2 != 0 && A[oddIndex] % 2 != 1) {
                temp = A[oddIndex]
                A[oddIndex] = A[evenIndex]
                A[evenIndex] = temp
            }
            if (A[evenIndex] % 2 == 0) {
                evenIndex += 2
            }
            if (A[oddIndex] % 2 == 1) {
                oddIndex += 2
            }
        }
        return A
    }
}

fun main(args: Array<String>) {
    val array = intArrayOf(4,2,5,7)
    val sortArrayByParityII = SortArrayByParityII()
    CommonUtils.printArray(sortArrayByParityII.sortArrayByParityII(array).toTypedArray())
}
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有問題隨時溝通

具體代碼實現能夠參考Github