leetcode 88 Merge Sorted Array

leetcode 88  Merge Sorted Array

题目描述：

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.

分析：
从后往前比较，将每次比较的较大值放入nums1中index所指的位置，这样就避免了反复的数据后移问题。
最坏情况下的时间复杂度O(n+m)。。。

C++代码：

class Solution {public:    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {        int index=m+n-1;        int index1=m-1;        int index2=n-1;        while(index1>=0 && index2>=0)        {            if(nums1[index1]<=nums2[index2])            {                nums1[index--]=nums2[index2--];            }            else            {                nums1[index--]=nums1[index1--];            }        }        while(index2>=0)        {            nums1[index--]=nums2[index2--];        }            }};

Python代码：

class Solution(object):    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.        """        index=m+n-1        index1=m-1        index2=n-1        while index1>=0 and index2>=0:            if nums1[index1]<=nums2[index2]:                nums1[index]=nums2[index2]                index-=1                index2-=1            else:                nums1[index]=nums1[index1]                index-=1                index1-=1        while index2>=0:            nums1[index]=nums2[index2]            index-=1            index2-=1