[LeetCode]Merge Sorted Array

题意;将A,B连个有序数组合并到A中,A空间充足思路1: 基本思路很简单,开辟一个额外数组,空间复杂度和时间复杂度都为O(N)代码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);}}思路2:从尾部反向遍历,将大的先放入数组的尾部空间复杂度为O(1),时间复杂度为O(N)代码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);}}}

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