88. Merge Sorted Array
Description
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
Note:
The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is equal to m + n) to hold additional elements from nums2.
Constraints
-10^9 <= nums1[i], nums2[i] <= 10^9
nums1.length == m + n
nums2.length == n
Approach
Links
GeeksforGeeks
YouTube
Examples
Input:
nums1 = [1, 2, 3, 0, 0, 0], m = 3
nums2 = [2, 5, 6], n = 3
Output: [1, 2, 2, 3, 5, 6]
Solutions
/**
* Time complexity : O(m + n)
* Space complexity : O(1)
*/
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int k = m+n-1;
while(m > 0 && n > 0) {
nums1[k--] = (nums1[m-1] > nums2[n-1])? nums1[--m]: nums2[--n];
}
while(n > 0) {
nums1[k--] = nums2[--n];
}
}
}/**
* Time complexity : O(m + n)
* Space complexity : O(1)
*/
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int k = m+n-1;
while(k >= 0) {
if(n < 1 || (m > 0 && nums1[m-1] > nums2[n-1])) {
nums1[k--] = nums1[--m];
} else {
nums1[k--] = nums2[--n];
}
}
}
}/**
* Time complexity : O(m * n)
* Space complexity : O(1)
*/
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
if(m > 0) {
for(int i = n-1; i >= 0; i--) {
int j, last = nums1[m-1];
for(j = m-2; j >= 0 && nums1[j] > nums2[i]; j--) {
nums1[j+1] = nums1[j];
}
if(j != m-2 || last > nums2[i]) {
nums1[j+1] = nums2[i];
nums2[i] = last;
}
}
}
for(int i = m; i < m+n; i++) {
nums1[i] = nums2[i-m];
}
}
}Follow up
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