LeetCode#88. Merge Sorted Array

88. Merge Sorted Array

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
 

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].
 

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
 

Constraints:

  • nums1.length == m + n
  • nums2.length == n
  • 0 <= m, n <= 200
  • 1 <= m + n <= 200
  • -109 <= nums1[i], nums2[j] <= 109

Follow up: Can you come up with an algorithm that runs in O(m + n) time?


直接合并,再冒泡排序

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        for (int i = m; i <m+n; i++) {
            nums1[i]=nums2[i-m];
        }
        for (int i = 0; i < m+n-1; i++) {
            for (int j = 0; j < m+n-i-1; j++) {
                if (nums1[j]>nums1[j+1]){
                    int tmp=nums1[j];
                    nums1[j]=nums1[j+1];
                    nums1[j+1]=tmp;
                }
            }
        }
    }
}

官方题解:直接合并,排序

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        for (int i = 0; i != n; ++i) {
            nums1[m + i] = nums2[i];
        }
        Arrays.sort(nums1);
    }
}
### LeetCode Problem 88 Python 解法 LeetCode88 题名为 **Merge Sorted Array**,其目标是将两个已排序数组合并成一个有序数组。以下是该问题的一个高效解决方案: #### 合并排序数组的核心逻辑 为了优化空间复杂度,通常采用从后向前填充的方法来解决此问题。这种方法避免了频繁移动元素的操作,从而提高了效率[^1]。 ```python class Solution: def merge(self, nums1, m, nums2, n): """ Do not return anything, modify nums1 in-place instead. """ p1, p2 = m - 1, n - 1 tail = m + n - 1 while p1 >= 0 and p2 >= 0: if nums1[p1] > nums2[p2]: nums1[tail] = nums1[p1] p1 -= 1 else: nums1[tail] = nums2[p2] p2 -= 1 tail -= 1 # 如果nums2还有剩余,则将其复制到nums1前面部分 nums1[:p2 + 1] = nums2[:p2 + 1] ``` 上述代码通过双指针技术实现了原地修改 `nums1` 的功能。它的时间复杂度为 \(O(m+n)\),其中 \(m\) 和 \(n\) 分别表示 `nums1` 和 `nums2` 中有效元素的数量。 --- #### 边界条件处理 需要注意的是,在某些情况下,如果 `nums1` 原始数据已经全部被覆盖完毕而 `nums2` 还有未处理的数据,则需要额外的步骤将这些数据拷贝至 `nums1` 的前部位置[^2]。 --- #### 测试用例验证 下面提供了一些测试用例以帮助理解算法的行为: ```python # 初始化对象实例 solution = Solution() # 定义输入参数 nums1 = [1, 2, 3, 0, 0, 0] m = 3 nums2 = [2, 5, 6] n = 3 # 调用方法执行操作 solution.merge(nums1, m, nums2, n) print(nums1) # 输出应为 [1, 2, 2, 3, 5, 6] ``` 以上代码展示了如何调用函数以及预期的结果形式[^3]。 ---
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值