Leetcode 每日一题C 语言版 -- 88 merge sorted array

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问题描述

复制代码

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?

解法一

clib qsort API

num2 的元素全贴到num1 后面n 个元素位置，然后对 nums1 使用qsort 排序

c 复制代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static int _compare(const void *a, const void *b)
{
    int __a = *(int *)a;
    int __b = *(int *)b;

    return (__a > __b) - (__a < __b);
}

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n) {

    memcpy((void *)(nums1 + m), (void *)nums2, sizeof(int) * n);
    qsort(nums1, m + n, sizeof(int), _compare);
}

解法2

nums1 后面有n 个空闲位置，考虑从后往前合并，避免了需memmove 数据迁移问题。

只需把num2 里面的数据全放到正确位置即可，剩下的nums1 由于已经是排好序的，自然在正确位置

c 复制代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n) {
    int nums1_index = m - 1;
    int nums2_index = n - 1;
    int pos = m + n - 1;


    while (nums2_index >= 0) {
        if ((nums1_index >= 0) && (nums1[nums1_index] > nums2[nums2_index])) {
            nums1[pos] = nums1[nums1_index];
            nums1_index--;
        } else {
            nums1[pos] = nums2[nums2_index];
            nums2_index--;
        }
        pos--;
    }
}