leetcode 中的合并排序数组

Question

我在https://leetcode.com/problems/merge-sorted-array/description/中看到了这个，答案在https://leetcode.com/problems/merge-sorted-array/discuss/29503/漂亮的 Python 解决方案

def merge(self, nums1, m, nums2, n):
        while m > 0 and n > 0:
            if nums1[m-1] >= nums2[n-1]:
                nums1[m+n-1] = nums1[m-1]
                m -= 1
            else:
                nums1[m+n-1] = nums2[n-1]
                n -= 1
        if n > 0:
            nums1[:n] = nums2[:n]

我有两个疑问：

1. 当我想运行代码时

   print(Solution().merge([1, 2, 3], 3,
   [4, 5, 6], 3))

   ，它只是给了我一个错误提示

   IndexError: list
   assignment index out of range

   。
2. 我不太明白上面答案中的方法论。 有人可以告诉我如何解决此类问题吗？

Answer 1

问题就在这里：

def merge(self, nums1, m, nums2, n):
    while m > 0 and n > 0:
        if nums1[m-1] >= nums2[n-1]:
            nums1[m+n-1] = nums1[m-1]   # nums1[m+n-1] IndexError
            m -= 1
        else:
            nums1[m+n-1] = nums2[n-1]   # Same as above
            n -= 1
    if n > 0:
        nums1[:n] = nums2[:n]

您可能需要先“调整”列表大小以防止索引错误：

def merge(self, nums1, m, nums2, n):
    nums1 += nums2
    # Or nums1 += [None] * len(nums2)

    while m > 0 and n > 0:
        if nums1[m-1] >= nums2[n-1]:
            nums1[m+n-1] = nums1[m-1] # nums1[m+n-1] IndexError
            m -= 1
        else:
            nums1[m+n-1] = nums2[n-1] # Same as above
            n -= 1
    if n > 0:
        nums1[:n] = nums2[:n]

Answer 2

def merge_two_sorted_arrays(left: list, right: list) -> list:
    """
    Merge Sorted Array in leetcode
    Args:
        left: left array
        right: right array

    Returns: A new merged array
    """
    A = [None] * (len(left) + len(right))
    i = j = k = 0
    while i < len(left) and j < len(right):
        if left[i] < right[j]:
            A[k] = left[i]
            i += 1
        else:
            A[k] = right[j]
            j += 1
        k += 1
    while i < len(left):
        A[k] = left[i]
        i += 1
        k += 1
    while j < len(right):
        A[k] = right[j]
        j += 1
        k += 1

    return A


a = [1, 2, 2, 4]
b = [3, 5, 6, 9]

print(merge_two_sorted_arrays(a, b))

输出：

[1, 2, 2, 3, 4, 5, 6, 9]

Answer 3

合并两个已排序数组的Java代码

给定两个已排序的数组，任务是以排序的方式合并它们。

输入：arr1[] = { 1, 3, 4, 5}，arr2[] = {2, 4, 6, 8} 输出：arr3[] = {1, 2, 3, 4, 4, 5, 6, 8}

输入：arr1[] = { 5, 8, 9}，arr2[] = {4, 7, 8} 输出：arr3[] = {4, 5, 7, 8, 8, 9}

尝试使用两个输入数组的不同组合并比较预期数组

导入java.util.Arrays；

公共类 MergeArraysDemo {

public static void main(String[] args) {
    int arr1[] = { 1, 3, 5, 7 };
    int arr2[] = { 2, 4, 6, 8 };
    // int arr1[] = { 5, 8, 9 };
    // int arr2[] = { 4, 7, 8 };
    int arr3[] = mergeArrays(arr1, arr2);
    System.out.println("Array after merging - " + Arrays.toString(arr3));
}

public static int[] mergeArrays(int[] arr1, int[] arr2) {
    int n = arr1.length;
    int m = arr2.length;

    int[] arr3 = new int[n + m];
    int j = 0;
    int k = 0;
    int min = 0;
    // max = Integer.MAX_VALUE;

    boolean flg = false;

    for (int i = 0; i < n + m; i++) {
        if (j < n && k < m) {
            min = arr2[k];
            if (arr1[j] <= min) {
                arr3[i] = arr1[j];
                j++;
                flg = false;
            } else {
                arr3[i] = min;
                k++;
                flg = true;
            }
        } else {
            if (flg) {
                arr3[i] = arr1[n - 1];
            } else {
                arr3[i] = min;
            }
        }
    }

    return arr3;
}

}