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[LeetCode 88] Merge Sorted Array
Problem
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
TypeScript
/**
Do not return anything, modify nums1 in-place instead.
*/
function merge(nums1: number[], m: number, nums2: number[], n: number): void {
// if nums2 is empty, we just return
if(n === 0) return;
let cur: number = m + n - 1;
let p1: number = m - 1;
let p2: number = n - 1;
// let's merge from back
while(p1 >= 0 && p2 >= 0){
if(nums1[p1] < nums2[p2]){
nums1[cur] = nums2[p2]
p2--;
} else {
nums1[cur] = nums1[p1];
p1--;
}
cur--;
}
// merge remaining element in nums2
while(p2 >= 0){
nums1[cur] = nums2[p2];
cur--;
p2--;
}
};