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MergedSortedArray.jav
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/**
*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.
*
*/
public class MergedSortedArray
{
class Solution
{
public void merge(int[] nums1, int m, int[] nums2, int n)
{
//get the last index of both arrays
int i = m - 1;
int j = n - 1;
//get the last index of combined array
int index = m + n - 1;
//loop through both arrays from the last value and compare them
while(j >= 0) //while nums2 still has uncheck values
{
//compare last values in nums1 and nums2
if( i >= 0 && nums1[i] > nums2[j])
{
//assign the greater value to last index in combined array
nums1[index] = nums1[i];
i--; //move to the next index
}
else
{
//thesame logic like in the if statement above
nums1[index] = nums2[j];
j--;
}
//move to the next index in combined array
index--;
}
}
}
}