[easy][Array][Two-pointer]88. Me

原题是：

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.

Screen Shot 2017-11-22 at 10.08.40 PM.png

思路是：

这个题，如果要不断比较两个数组的元素，然后插入，需要挪动大量元素的位置，且是多次，时间复杂度太大。
那么我们可以借助第一个数组后面“空”出来的空间，从后向前依次“排队”。

代码是：

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.
        """
        i,j = m-1,n-1
        cnt = 1
        while i > -1 and j > -1 :
            if nums1[i] > nums2[j]:
                nums1[m + n - cnt] = nums1[i]
                i -= 1
            else:
                nums1[m + n - cnt] = nums2[j]
                j -= 1
            cnt += 1
        
        if i == -1:
            for p in range(0,j+1):
                nums1[p] = nums2[p]