Question

Input : arr = {-1, -1, 6, 1, 9, 3, 2, -1, 4, -1}
Output : [-1, 1, 2, 3, 4, -1, 6, -1, -1, 9]

Input : arr = {19, 7, 0, 3, 18, 15, 12, 6, 1, 8,
              11, 10, 9, 5, 13, 16, 2, 14, 17, 4}
Output : [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 
         11, 12, 13, 14, 15, 16, 17, 18, 19]

方法 1.导航阵列。

检查a [i] = -1，如果是，则忽略它。
如果a [i]！= -1，检查元素a [i]是否在其正确位置（i = A [i]）。如果是，则忽略它。
如果a [i]！= -1且元素a [i]不在其正确位置（i！= A [i]），则将其置于正确位置，但有两个条件：

（i）.A [i]是空的，意味着A [i] = -1，然后只是把A [i] = i。

（ii）.OR A [i]不是空的，意味着A [i] = x，则int y = x put A [i] = i。现在，我们需要将y放在正确的位置，重复步骤3。

以下解决方案的时间复杂度是多少？

public static int[] fix(int[] A) {
    for (int i = 0, x = A[i]; i < A.length; i++) {
        if (x == -1 || x == i)
            continue;

        // check if desired place is not vacant
        while (A[x] != -1 && A[x] != x) {
            int y = A[x];   // store the value from desired place
            A[x] = x;       // place the x to its correct position
            x = y;          // now y will become x, now search the place for x
        }

        A[x] = x;           // place the x to its correct position

        // check if while loop hasn't set the correct value at A[i]
        if (A[i] != i)
            A[i] = -1;      // if not then put -1 at the vacated place
    }

    return A;
}

Answer 1

我可以为您提供两种算法。

第一个：使用额外数组，时间复杂度为O(n)，O(n)额外内存

public static int[] fix(int[] arr) {
    int[] res = new int[arr.length];

    Arrays.fill(res, -1);

    for (int i = 0; i < arr.length; i++)
        if (arr[i] != -1)
            res[arr[i]] = arr[i];

    return res;
}

第二个：到位，在最坏的情况下，时间复杂度为O(n^2)，在平均情况下为O(n)，没有额外的内存

public static int[] fix(int[] arr) {
    int i;

    while ((i = findIncorrectPosition(arr)) >= 0) {
        while (arr[i] != -1 && arr[i] != i) {
            swap(arr, i, arr[i]);
        }
    }

    return arr;
}

...加上两种私人支持方式：

private static int findIncorrectPosition(int[] arr) {
    for (int i = 0; i < arr.length; i++)
        if (arr[i] != -1 && arr[i] != i)
            return i;
    return -1;
}

private static void swap(int[] arr, int i, int j) {
    int tmp = arr[i];
    arr[i] = arr[j];
    arr[j] = tmp;
}

Answer 2

f(n) = O(n) 使用交换方法#Swift

for i in stride(from: 0, to: array.count, by: 1){
      if(array[i] >= 0 && array[i] != i){
        let ele = array[array[i]]
        array[array[i]] = array[i]
        array[i] = ele
       
      }
    }
    
    for k in stride(from: 0, to: array.count, by:1 ){
      print("  ",array[k])
    }

重新排列一个数组，使arr [i] = i

2 个答案: