88. Merge Sorted Array
題目敘述
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 + nnums2.length == n0 <= m, n <= 2001 <= 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?
題目翻譯
給定兩個整數陣列 nums1 和 nums2，其中這兩個陣列的元素是以升冪排列。另外分別在給兩個整數參數 m 和 n，代表分別從 nums1 和 nums2 取出的元素數量。將 m 和 n 個元素統一編輯到 nums1 中，並且依照升冪排序。不需要回傳值。所以最後 nums1 應該會要 m + n 的長度。
另外思考看看如何將時間複雜度做到 O(m + n)
解法解析
這題有三種解法，第一種就是很直覺的方式，將兩者先相加後再做排序。這種方式就基本上依靠程式語言內建支援的函數來做排序處理。
Time complexity：O((n+m)log(n+m))
Space complexity：O(n)
剩下兩種方式，因為兩著陣列是已經排序過的，所以利用此條件下，使用 Three Pointers 的解法，其差異只是分別從頭開始遍歷還是從尾部開始遍歷。
Three Pointers 分別指向：m 的數量、n 的數量、要插入 nums1 目前的索引位置。這兩種方式都能達到時間複雜度 O(m+n)，但是從頭開始的話，空間複雜度會是 O(m)，從尾部的空間複雜度則是 O(1)。
解法範例
Go
Merge then sort
func merge(nums1 []int, m int, nums2 []int, n int) {
for i := 0; i < n; i++ {
nums1[m+i] = nums2[i]
}
sort.Ints(nums1)
}
Three Pointers (Start From the Beginning)
func merge(nums1 []int, m int, nums2 []int, n int) {
nums1Copy := make([]int, m)
copy(nums1Copy, nums1)
var p1, p2 int = 0, 0
for p := 0; p < n+m; p++ {
if p2 >= n || (p1 < m && nums1Copy[p1] <= nums2[p2]) {
nums1[p] = nums1Copy[p1]
p1++
} else {
nums1[p] = nums2[p2]
p2++
}
}
}
Three Pointers (Start From the End)
func merge(nums1 []int, m int, nums2 []int, n int) {
var p1, p2 int = m - 1, n - 1
for i := n + m - 1; i >= 0; i-- {
if p2 < 0 {
break
}
if p1 >= 0 && nums1[p1] > nums2[p2] {
nums1[i] = nums1[p1]
p1--
} else {
nums1[i] = nums2[p2]
p2--
}
}
}
JavaScript
Merge then sort
/**
* @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) {
for (let i = 0; i < n; i++) {
nums1[i + m] = nums2[i];
}
return nums1.sort((a, b) => a - b);
};
Three Pointers (Start From the Beginning)
/**
* @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) {
const nums1Copy = nums1.slice(0, m);
let p1 = 0, p2 = 0;
for (let i = 0; i < m + n; i++) {
if (p2 >= n || (p1 < m && nums1Copy[p1] <= nums2[p2])) {
nums1[i] = nums1Copy[p1];
p1++;
} else {
nums1[i] = nums2[p2];
p2++;
}
}
};
Three Pointers (Start From the End)
/**
* @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 p1 = m - 1,
p2 = n - 1;
for (let i = n + m - 1; i >= 0; i--) {
if (p2 < 0) {
break;
}
if (p1 >= 0 && nums1[p1] > nums2[p2]) {
nums1[i] = nums1[p1];
p1--;
} else {
nums1[i] = nums2[p2];
p2--;
}
}
};
Kotlin
PHP
Python
Merge then sort
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
# Write the elements of num2 into the end of nums1.
for i in range(n):
nums1[i + m] = nums2[i]
# Sort nums1 list in-place.
nums1.sort()
Three Pointers (Start From the Beginning)
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
# Make a copy of the first m elements of nums1.
nums1_copy = nums1[:m]
# Read pointers for nums1Copy and nums2 respectively.
p1 = 0
p2 = 0
# Compare elements from nums1Copy and nums2 and write the smallest to nums1.
for p in range(n + m):
# We also need to ensure that p1 and p2 aren't over the boundaries
# of their respective arrays.
if p2 >= n or (p1 < m and nums1_copy[p1] <= nums2[p2]):
nums1[p] = nums1_copy[p1]
p1 += 1
else:
nums1[p] = nums2[p2]
p2 += 1
Three Pointers (Start From the End)
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
# Set p1 and p2 to point to the end of their respective arrays.
p1 = m - 1
p2 = n - 1
# And move p backwards through the array, each time writing
# the smallest value pointed at by p1 or p2.
for p in range(n + m - 1, -1, -1):
if p2 < 0:
break
if p1 >= 0 and nums1[p1] > nums2[p2]:
nums1[p] = nums1[p1]
p1 -= 1
else:
nums1[p] = nums2[p2]
p2 -= 1
Rust
Swift