[LeetCode] 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?

合并两个有序数组。

给你两个按非递减顺序排列的整数数组 nums1 和 nums2，另有两个整数 m 和 n ，分别表示 nums1 和 nums2 中的元素数目。

请你合并 nums2 到 nums1 中，使合并后的数组同样按非递减顺序排列。

注意：最终，合并后数组不应由函数返回，而是存储在数组 nums1 中。为了应对这种情况，nums1 的初始长度为 m + n，其中前 m 个元素表示应合并的元素，后 n 个元素为 0 ，应忽略。nums2 的长度为 n 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/merge-sorted-array
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既然两个数组都已经排好序，而且nums1的size保证能塞得下两个数组长度之和。思路是从后往前比较谁的元素大，谁就把元素放到nums1里面。这样nums1会被下标从右到左，值从大到小被填满。

时间O(m + n)

空间O(1)

Java实现

 1 class Solution {
 2     public void merge(int[] nums1, int m, int[] nums2, int n) {
 3         m = m - 1;
 4         n = n - 1;
 5         int i = m + n + 1;
 6         while (m >= 0 || n >= 0) {
 7             if (m < 0) {
 8                 nums1[i] = nums2[n];
 9                 i--;
10                 n--;
11             } else if (n < 0) {
12                 nums1[i] = nums1[m];
13                 i--;
14                 m--;
15             } else {
16                 nums1[i--] = nums1[m] > nums2[n] ? nums1[m--] : nums2[n--];
17             }
18         }
19     }
20 }

JavaScript实现

 1 /**
 2  * @param {number[]} nums1
 3  * @param {number} m
 4  * @param {number[]} nums2
 5  * @param {number} n
 6  * @return {void} Do not return anything, modify nums1 in-place instead.
 7  */
 8 var merge = function(nums1, m, nums2, n) {
 9     m = m - 1;
10     n = n - 1;
11     let i = m + n + 1;
12     while (m >= 0 || n >= 0) {
13         if (m < 0) {
14             nums1[i] = nums2[n];
15             i--;
16             n--;
17         } else if (n < 0) {
18             nums1[i] = nums1[m];
19             i--;
20             m--;
21         } else {
22             nums1[i--] = nums1[m] > nums2[n] ? nums1[m--] : nums2[n--];
23         }
24     }
25 };

LeetCode 题目总结