Question

我试图将所有非零整数移动到数组的开头/左侧＆＃39; a。＆＃39;（例如{0,0,4,3,0}变为{4 ，3,0,0,0}）

这是我的计划。它编译并运行没有错误，但数组最终只有零。任何建议将不胜感激。

int[] squeezeLeft(int[] a) {
    int count = 0;
    //creates new array
    int [] a2 = new int[a.length];
    //loops through a
    for (int i = 0; i< a.length; i++){
        //gets a temporary value from a[i]
        int temp = a[i];
        //assigns a[i] to temporary variable
        a[count] = temp;
        a[i] = 0;
                  /* raises count integer by one, so the following indice which is zero, is replaced

                  by  the following non=zero integer*/
        count++;
    }
    return a2;
}

Answer 1

我知道这不是一个非常有效的解决方案O^2，但它会按照你的要求行事。

private static int[] sequeezeLeft(final int[] a) {
    for (int i = 0; i < a.length; i++) {
        if (a[i] != 0) {
            for (int j = 0; j < a.length; j++) {
                if (a[j] == 0) {
                    a[j] = a[i];
                    a[i] = 0;
                }
            }
        }
    }
    return a;
}

O(n)时间复杂度的另一个版本

    private static int[] sqeeze2(int [] a){
    int index = 0;
    if(a.length == 0)
        return a;
    for(int i=0;i<a.length;i++){
        if(a[i] !=0 && index < a.length){
            a[index] = a[i];
            a[i]=0;
            index++;
        }
    }
    return a;
}

Answer 2

或者如果你有点懒，那么使用它呢？

    Integer[] ints = new Integer[] { 0, 5, 2, 0, 1, 5, 6 };
    List<Integer> list = Arrays.asList(ints);
    Collections.sort(list);
    Collections.reverse(list);
    System.err.println(list); // prints [6, 5, 5, 2, 1, 0, 0]

不确定性能，但是..

Answer 3

static void squeezeLeft(int[] array) {
    int arrayIndex = 0;
    if (array.length != 0) {//check the length of array whether it is zero or not

        for (int i = 0; i < array.length; i++) {
            if (array[i] != 0 && arrayIndex < array.length) {//check the non zero element of array
                if (i != arrayIndex) {//if the index of non zero element not equal to array index 
                                      //only then swap the zero element and non zero element
                    array[arrayIndex] = array[i];
                    array[i] = 0;
                }
                arrayIndex++;  //increase array index after finding non zero element
            }
        }
    }
}

Answer 4

与输入1,0,2,0,4,0,5,0,6,0一样，第二个解决方案将失败，因为输出将是：

0245600000

对于o（n）：

private static int[] reArrangeNonZeroElement(int arr[]) {
    int count = 0;
    if (arr.length == 0)
        return arr;
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] != 0) {
            arr[count++] = arr[i];
        }
    }

    for (; arr.length > count; ) {
        arr[count++] = 0;
    }
    return arr;
}

Answer 5

怎么样：

Arrays.sort(ints, new Comparator<Integer>() {
    @Override
    public int compare(Integer o1, Integer o2)
    {
        if (o1 == 0) return 1;
        if (o2 != 0 && o1 > o2) return 1;
        if (o2 != 0 && o1 == o2) return 0;
        return -1;
    }
});

将所有非零整数移动到数组的开头

5 个答案: