88. 合并两个有序数组
88. 合并两个有序数组
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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