88 Merge Sorted Array

Leetcode 88

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.

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        // base case
        if (m == 0 && n == 0){
            return;
        }
        if (m==0 && n !=0){
            for (int i = 0; i < n; i++){
                nums1[i] = nums2[i];
            }
            return;
        }

        if (m != 0 && n == 0){
            return;
        }

        // [1 2, 3, 0, 0,0 ]
        //       i
        //               k
        // [2, 5, 6]
        //        j

        int i = m -1;               // i 代表了 数量
        int j = n -1;               // j 也是数量
        int k = nums1.length -1;        // 三个指针 不要搞错了
        while(i >=0 && j >=0){
            if (nums1[i] < nums2[j]){
                nums1[k] = nums2[j];
                j--;
                k--;
            }else{
                nums1[k] = nums1[i];
                if (k != i){
                    nums1[i] = 0;
                }
                i--;
                k--;
            }
        }

        while(j >= 0){
            nums1[k] = nums2[j];
            k--;
            j--;
        }
    }
}

// TC: O(n)
// SC: O(1)

/* Leetcode 88 Merge Sorted Array
 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.
*/
import java.util.*;
public class MergeSortedArray88{
    public static void main(String[] args){
        int[] nums1 = new int[]{1,2,3,0,0,0};
        int[] nums2 = new int[]{2,5,6};
        int m = 3;
        int n = 3;

        int[] a2 = new int[]{1};
        int[] b2 = new int[]{};
        int a = 1;
        int b = 0;

        int[] a3 = new int[]{0};
        int[] b3 = new int[]{1};
        int x = 0;
        int y = 1;

        System.out.println(Arrays.toString(nums1));
        System.out.println(Arrays.toString(a2));
        System.out.println(Arrays.toString(a3));

        Merge(nums1, m, nums2, n);
        Merge(a2, a, b2, b);
        Merge(a3,x,b3,y);
        System.out.println(Arrays.toString(nums1));
        System.out.println(Arrays.toString(a2));
        System.out.println(Arrays.toString(a3));


    }


    public static void Merge(int[] nums1, int m, int[] nums2, int n){
        /*
         *1 2 3 0 0 0 
              p1    i

                2 5 6
                p2
        */

        int p1 = m-1;
        int p2 = n-1;

        for (int i = m + n -1; i >=0 ; i--){
            if (p2 < 0){
                break;
            }

            if (nums1[i] < nums2[p2]){
                nums1[i] = nums2[p2];
                p2--;
            }else{ // nums1[i] >= nums[p2]
                nums1[i] = nums1[p1];
                p1--;
            }
        }


    }

    /*
     * Time complexity: O(n)
     * Space: O(1)
     */

}