Merge Sorted Array

test instructions; Merge A, B with an ordered array into a, a space sufficient
Idea 1: The basic idea is simple, open up an extra array, space complexity and time complexity are O (N)
Code Listing 1:

public class Solution {public    void merge (int a[], int m, int b[], int n) {        if (n = = 0) return;        if (M = = 0) {            system.arraycopy (B, 0, A, 0, N);            return;        }        int i =0, j=0, temp = 0,k=0;        int [] C = new Int[a.length];        while (I < m && J < N) {            while (I < M && A[i] <= b[j]) {                c[k + +] = a[i + +];            }            while (I < m && J < n && B[j] < A[i]) {                c[k + +] = b[j + +];            }        }        if (I < m) {            system.arraycopy (A, I, C, K, m-i);        }        if (J < n) {            system.arraycopy (B, J, C, K, n-j);        }        System.arraycopy (C, 0, A, 0, c.length);}    }

Idea 2: Reverse traversal from the tail, the large first into the array of the trailing space complexity of O (1), the time complexity of O (N)
Code Listing 2:

public class Solution {public    void merge (int a[], int m, int b[], int n) {        if (n = = 0) return;        if (M = = 0) {            system.arraycopy (B, 0, A, 0, N);            return;        }        int tail = m + n-1;        n--;m--;        while (M >= 0 && n >=0) {            A[tail--] = Math.max (A[m], b[n]);            if (A[m] >= b[n]) m--;            else n--;        }        while (n--> 0) {        //    a[tail--] = B[n];        }        if (n >= 0) {            system.arraycopy (b,0,a,0,n+1);}}    }

Merge Sorted Array