88. 合并两个有序数组

🟢   🔖  数组 双指针 排序  🔗 力扣 LeetCode

题目

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 arraynums1. 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
• -10^9 <= nums1[i], nums2[j] <= 10^9

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

题目大意

给你两个按 递增顺序 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 递增顺序 排列。

解题思路

为了不大量移动元素，就要从 2 个数组长度之和的最后一个位置开始，依次选取两个数组中大的数，从第一个数组的尾巴开始往头放，只要循环一次以后，就生成了合并以后的数组了。

复杂度分析

• 时间复杂度：O(n + m)，其中m 和 n ，分别表示 nums1 和 nums2 中的元素数目，需要遍历一遍两个数组。
• 空间复杂度：O(1)，用了常数个变量存储中间状态。

代码

/**
 * @param {number[]} nums1
 * @param {number} m
 * @param {number[]} nums2
 * @param {number} n
 * @return {void} Do not return anything, modify nums1 in-place instead.
 */
var merge = function (nums1, m, nums2, n) {
	let i = m - 1;
	let j = n - 1;
	let k = m + n - 1;
	while (j >= 0) {
		if (nums1[i] > nums2[j]) {
			nums1[k] = nums1[i];
			i--;
		} else {
			nums1[k] = nums2[j];
			j--;
		}
		k--;
	}
};

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