Java&Go-LeetCode922. 按奇偶排序数组 II-双指针

• 算法
  • 1.双指针
  • 2.一个指针在奇数位置上移动，一个指针在偶数位置上移动
  • 3.每次移动到奇数位置上是偶数，偶数位置上是奇数，交换二者，继续移动

public int[] sortArrayByParityII(int[] A) {
    int n = A.length;
    int i = 0, j = 1;
    while (i < n && j < n) {
        while (i < n && (A[i] & 1) == 0) {
            i += 2;
        }
        while (j < n && (A[j] & 1) == 1) {
            j += 2;
        }
        if (i < n && j < n) {
            swap(A, i, j);
        }
        i += 2;
        j += 2;
    }
    return A;
}

private void swap(int[] A, int x, int y) {
    int temp = A[x];
    A[x] = A[y];
    A[y] = temp;
}

func sortArrayByParityII(A []int) []int {
    n := len(A)
    i, j := 0, 1
    for i < n && j < n {
        for i < n && (A[i] & 1) == 0 {
            i += 2
        }
        for j < n && (A[j] & 1) == 1 {
            j += 2
        }
        if i < n && j < n {
            A[i], A[j] = A[j], A[i]
        }
        i, j = i + 2, j + 2
    }
    return A
}

• 算法
  • 1.奇数位置上放上奇数时，偶数位置上自动放好偶数
  • 2.每次循环，一直和偶数位置上的数交换，直到该奇数位置上奇数为止

public int[] sortArrayByParityII(int[] A) {
    int n = A.length;
    int even = 0, odd = 1;
    for (; even < n && odd < n; odd += 2) {
        while ((A[odd] & 1) == 0) {
            int temp = A[even];
            A[even] = A[odd];
            A[odd] = temp;
            even += 2;
        }
    }
    return A;
}

func sortArrayByParityII(A []int) []int {
    n := len(A)
    even, odd := 0, 1
    for ; even < n && odd < n; odd += 2 {
        for (A[odd] & 1) == 0 {
            A[even], A[odd] = A[odd], A[even]
            even += 2
        }
    }
    return A
}