LeetCode【88】 Merge Sorted Array

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

/** * Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. * 将数组2merge到数组1中，数组1有足够的长度 * * @auther Dennis * @date 2017/12/12 * 思想：从num1的尾部开始往前进行插入，这样就不用每个元素都移动了,时间复杂度O(m+n),空间复杂度O(1) */public class MergeSortedArray {    public static void merge(int A[], int m, int B[], int n) {        if (n == 0) {            return;        }        if (m == 0 && n != 0) {            for (int i = 0; i < n; i++) {                A[i] = B[i];            }        }        int i = m - 1;        int j = n - 1;        int k = m + n - 1;        while (i >= 0 && j >= 0) {    //分别从数A和B最后的元素开始向前比较            if (A[i] > B[j]) {                A[k] = A[i];                i--;            } else {                A[k] = B[j];                j--;            }            k--;        }        while (i >= 0) {            A[k] = A[i];            i--;            k--;        }        while (j >= 0) {            A[k] = B[j];            j--;            k--;        }    }    public static void main(String[] args) {        int[] A = new int[]{3, 4, 6, 8, 9, 0, 0, 0, 0};        int[] B = new int[]{2, 4, 5, 7};        int m = 5;        int n = 4;        merge(A, m, B, n);        for (int i = 0; i < A.length; i++) {            System.out.print(A[i] + " ");        }    }}