[LeetCode] 88. Merge Sorted Array 合并有序数组

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

Note:

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

Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

给两个有序的数组 A 和 B，把 B 合并到 A，变成一个数组，假定 A 有足够的空间。

可以考虑新建立一个m + n的数组，然后从两个数组的开头取各取一个元素进行比较，把小的放进新数组，然后在循环这个过程，直到结束。

好的方法是不用新建立数组，而是直接在A 数组上写入，因为 A 足够大，可从两个数组的最大数也就是最后一个数开始比较，大的写入A[m + n -1]，然后循环这个过程。如果 B 的元素写完了，A 剩下的元素正好在正取的位置，不用写了。如果 A 的元素都取完了，那剩下的 B 的元素可一次全部写进 A。

Java:

public class Solution {
    public void merge(int A[], int m, int B[], int n) {
  
        while(m > 0 && n > 0){
            if(A[m-1] > B[n-1]){
                A[m+n-1] = A[m-1];
                m--;
            }else{
                A[m+n-1] = B[n-1];
                n--;
            }
        }
  
        while(n > 0){
            A[m+n-1] = B[n-1];
            n--;
        }
    }
}

Java:

public void merge(int A[], int m, int B[], int n) {
    int i = m - 1;
    int j = n - 1;
    int k = m + n - 1;
  
    while (k >= 0) {
        if (j < 0 || (i >= 0 && A[i] > B[j]))
            A[k--] = A[i--];
        else
            A[k--] = B[j--];
    }
}

Python:

class Solution:
    # @param A  a list of integers
    # @param m  an integer, length of A
    # @param B  a list of integers
    # @param n  an integer, length of B
    # @return nothing
    def merge(self, A, m, B, n):
        last, i, j = m + n - 1, m - 1, n - 1
 
        while i >= 0 and j >= 0:
            if A[i] > B[j]:
                A[last] = A[i]
                last, i = last - 1, i - 1
            else:
                A[last] = B[j]
                last, j = last - 1, j - 1
 
        while j >= 0:
                A[last] = B[j]
                last, j = last - 1, j - 1
 
if __name__ == "__main__":
    A = [1, 3, 5, 0, 0, 0, 0]
    B = [2, 4, 6, 7]
    Solution().merge(A, 3, B, 4)
    print(A)

Python:

class Solution:
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        while m > 0 and n > 0:
            if nums1[m-1] > nums2[n-1]:
                nums1[m+n-1] = nums1[m-1]
                m -= 1
            else:
                nums1[m+n-1] = nums2[n-1]
                n -= 1
        if n > 0:
            nums1[:n] = nums2[:n]

C++:

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i = m + n;
        while (m > 0 && n > 0) {
            if (nums1[m - 1] > nums2[n - 1]) {
                nums1[i - 1] = nums1[m - 1];
                --m;
            } else {
                nums1[i - 1] = nums2[n - 1];
                --n;
            }
            --i;
        }
 
        while (n > 0) {
            nums1[i - 1] = nums2[n - 1];
            --n;
            --i;
        }
    }
};