# 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 + 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?

## 解法解析

Time complexity：`O((n+m)log(n+m))` Space complexity：`O(n)`

### 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

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#### PHP

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#### 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

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#### Swift

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